Offer

2018年长沙理工大学第十三届程序设计竞赛题解

链接:https://www.nowcoder.com/acm/contest/96/A
来源:牛客网

“LL是什么?这都不知道的话,别说自己是程序猿啊!”
  “longlong?”
  “。。。肯定是LoveLive啊!”
  qwb为了检验你是否是真正的程序猿,决定出道题考考你:现在程序会输入一行字符串,如果恰好是lovelive(不区分大小写)就输出yes,否则输出no。

输入描述:

输入有多组(组数不超过100),每组输入一行字符串(字符串长度不超过100)。

输出描述:

输出占一行,如果输入符合要求则输出yes,否则输出no。

示例1

输入

longlong
LoveLive
love live

输出

no
yes
no

分析:主要是想了解一下C++中tolower语句

  1 /**
  2  
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 69                                                                     :r7Ji;r7;::;;.
 70                                                                         .                                                                             **/
 71                                                                         
 72 #include <bits/stdc++.h>
 73 using namespace std;
 74 
 75 #define FF(i, a, b) for(int i = a; i < b; i++)
 76 #define RR(i, a, b) for(int i = a; i > b; i++)
 77 #define ME(a, b) memset(a, b, sizeof(a))
 78 #define SC(x) scanf("%d", &x)
 79 #define PR(x) printf("%d\n", x)
 80 #define INF 0x3f3f3f3f
 81 #define MAX 1001
 82 #define MOD 1e9+7
 83 #define E 2.71828182845
 84 #define PI 3.1415926
 85 #define M 8
 86 #define N 6
 87 typedef long long LL;
 88 typedef pair<int, int> Author;
 89 vector<pair<string, int> > VP; 
 90 
 91 int main(void){
 92     ios::sync_with_stdio(false); cin.tie(0);
 93 //  freopen("in.txt", "r", stdin);
 94 //  freopen("out.txt", "w", stdout);
 95     int i, len1;string str = "lovelive", str1;
 96     
 97     while(getline(cin, str1)){
 98         len1 = str1.size();
 99         for(i = 0; i < len1; i++)
100             str1[i] = tolower(str1[i]);
101         if(str1 != str) cout<<"no"<<endl;
102         else cout<<"yes"<<endl;
103     }
104 
105 
106     return EXIT_SUCCESS;
107 }
View Code

链接:https://www.nowcoder.com/acm/contest/96/C
来源:牛客网

题目描述

durong有a台iphonex和b台s8,并且放在一个保险箱里,durong现在一台一台从保险箱随机拿出这些手机,现在他想知道第k次拿出s8的概率是多少

输入描述:

第一行一个正整数T,表示数据组数。(1<=T<=10000)
接下来T行输入a,b,k
其中(1<=a,b,k<=1e9) k<=a+b;

输出描述:

第k次拿出s8的概率,保留三位小数
示例1

输入

1
1 1 1

输出

0.500

分析:前面拿球不影响后面,所以第k次拿球是C(b, 1),总排列组合是A(a + b, a + b),拿去之后是A(a + b -1 , a + b - 1);

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 #define FF(i, a, b) for(int i = a; i < b; i++)
 5 #define RR(i, a, b) for(int i = a; i > b; i++)
 6 #define ME(a, b) memset(a, b, sizeof(a))
 7 #define SC(x) scanf("%d", &x)
 8 #define PR(x) printf("%d\n", x)
 9 #define MAX 1000
10 #define M 8
11 #define N 6
12 typedef long long LL;
13 typedef pair<int, int> Author;
14 
15 int main(void){
16     int T, a, b, k;
17     
18     cin>>T;
19     while(T--){
20         cin>>a>>b>>k;
21         cout<<fixed<<setprecision(3)<<b * 1.0 / (a + b) * 1.0<<endl;
22     }
23 
24     return 0;
25 }
View Code

链接:https://www.nowcoder.com/acm/contest/96/D
来源:牛客网

时间限制:C/C++ 1秒,其他语言2秒
空间限制:C/C++ 32768K,其他语言65536K
64bit IO Format: %lld

题目描述

    离散数学中有种名叫“哈斯图”的东西。

    在这题中,你们需要计算的是一些正整数在偏序关系“整除”下的哈斯图的边数。用大白话讲,在偏序关系“整除”下的哈斯图,就是把一个个正整数看成一个个图的节点,某些节点之间有边。连边的规则是这样的:对于任意两个正整数a和b(a<b)来说,如果b%a==0,并且不存在一个正整数c(a<c<b),使得条件b%c==0和c%a==0同时成立,那么我们就在节点a和节点b之间连一条边。

    现在问题是,给你们2个数L,R(1<=L,R<=1e6)。求由L,L+1,L+2...R这R-L+1个正整数在偏序关系“整除”下的哈斯图的边数。

    比如L=1,R=4,节点的组合有(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)。组合(1,2),(1,3),(2,4)可以连边。(1,4)因为中间存在c=2,不符合连边条件。所以当L=1,R=4的时候,这个哈斯图有3条边。

输入描述:

多组输入,不超过1000组数据
每组数据一行,包含2个正整数L和R,中间由空格分开。

输出描述:

每组数据输出一行,包含一个整数表示哈斯图的边数。
示例1

输入

1 4
4 10
1 10

输出

3
2
11

分析:就是a,c中间不能有b能被C, a整除,筛选素数,判断p[0]是为了控制边界,再计算p[i]中的素数即可

  1 /**
  2  
  3                                                                        :;LaEaHKEEGpPXU7;,
  4                                                                   .:75pKH11252U252XapZgRQgD6XJscLr;,.
  5                                                                :LXpRgGaX521JLw1JswJJsJs22XHPPEZEGDOMMRDOa7.
  6                                                            .r2EDDZEpZPZP6KpHX5SXH5XXa5KwaXaSX5UJ1c77sLs2GMQQ6r                                       .
  7                                                         ,LpgOGpEZGZEZEpZKpHHU5wP5HEDgpXpHa2SSa5aSXULr7rrirrJXRBp;                                   ;B
  8                                                      ,J6MRZH6EgEEZE6E6EZZPZXXwSSGQXr::aPpP5USUHaHaKa5Lvrr7ri;rLHBB2:                                Kc
  9                                                    rpQDOpPPOGGZOGOZG6GEOEOEDPPGBa.  .PaSSUXSUUUaUSaKXKS177r7rrrirSBBR7                             .O,
 10                                                 :UBQOKPK6ZOOOEDEO6GZE6EpEpDgDBR:   UBpXHa5aSaUS5SUS5XapPHJc7rrv7rr7sgBBs                           .g.
 11                                               ;gBMPXpO6GEOEOEOEGEOEE6EZEEDRGBB    EB5pKSXpHKaHSX552S5aUHHEX17c7vr7777s5RBS:                        .R;
 12                                            .sQBPXpDZOODOgODGOGgEDEOEOGgGgOOMB:   LBKKSXSHa52aaKXHXKaa5aSaaHXSJLcL7vcc777JDBBg2;.                    Qi
 13                                          ;2ggp2EDDOggGEDGgDDOgGGZDOOZOGg6gEBX    vBZaHUKaaUXXXSXXKXpXHXH5wwaa52U1wssLsLJccv1UDQBQ67.                O7
 14                                        :ZZUU5PROOEOZOZGGOODZOGgODZOOgOggRgRB; ;:..R6XaKKpP6PGppKPHpHpPX5aU21UUa5Sw52UwUJJv77L77sSpQMDU;            ;B5
 15                                      ,SRJ7sSHGggEOZOEG6OODpOZgggOQQBBBBBQBBQ.,;;. LBOgOgDRDDZODMgQRgRDaa552a252UUa25w5UaU2sLvccs7r7sJZBBMr       ,XQJ:
 16                                     LQHr77J6RGOZDZOEGEDGgDRORQBBBQRDPU1Jscwa7.,::.:J7r;::::.  ..,:;i7UOgRRRgDPH5SUSU52U2HHa1JJJLJLccLr71RBB,    7R2,
 17                                   :RZv;77JSgGOZEODEDGOEggQBBBMS7;:,:,,.,.,.:7L;:,;.:  ,: . .....       .:;rJU6GgGRggEZHPaKXX2S221Js1Lc7r:7QB. .XX:
 18                                  7g1;;7rcXG6gpGDZGgZOOQBBQpr.   ::::;:;:::Jr::sr;::;:.:vs,:::::::::::,,         ,:7L5HGOggRgZUUU5wSUaJLc7r7BOiDr  ...
 19                                .XX;;;irLHGKpZZZEKgDRBBB6i    ,;;:;;;;r;;:s177:,;L7:;7:.rHi,:::::::::::::::::,.         .,;7ZRQgO6KUUJsLwsJ7KBM. .....
 20                                JZ:;rrrc5EPHp6XgpRBBBE;     :i:;;;;;:;;;:c7::r7;,::::::::rv:,:::::::::::,:,:,:,,,,            .;sORQRGX21wsXU:  .... .
 21                               .Br;iir72EHPHZ6EgBBR7.   ..;ii:;;;;;;;;;:71r::.:7,  .::7;:;H;.::,:j::::::,:::::,,,::,,,              ,7wEDRZBMr  .......
 22                               1D;:r;rwOPXPKH6BBX,   .:;:;;;:::;:;;;;;::Ls,;ss..r.  ,c77;sLU,:,:::,:::,:,:,:,,,:,,,:::.                  .:.rP:.......
 23                               D2:i;rJpKHXHXgg7    .,,::::;:::::;:;:;;:;SL7sS2, :.   ::::,:U7.:::,:::,:,,,,,:,,.,,,.,:,                    ;L:. .. ...
 24                              :Qc;i7LGZPa6gBM,  ...,,::::::::::::;;::;.JJ;ic:         ;:::,v1,.:::;7,,,:,:,:,,,,.:,,,,..                :2wr.  .......
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 68                                                                   .:J: :;  .,:K6BQS7:.,.,.
 69                                                                     :r7Ji;r7;::;;.
 70                                                                         .                                                                             **/
 71                                                                         
 72 #include <bits/stdc++.h>
 73 using namespace std;
 74 
 75 #define FF(i, a, b) for(int i = a; i < b; i++)
 76 #define RR(i, a, b) for(int i = a; i > b; i++)
 77 #define ME(a, b) memset(a, b, sizeof(a))
 78 #define SC(x) scanf("%d", &x)
 79 #define PR(x) printf("%d\n", x)
 80 #define INF 0x3f3f3f3f
 81 #define MAX 1000005
 82 #define MOD 1e9+7
 83 #define E 2.71828182845
 84 #define PI 3.14159265358979
 85 #define M 8
 86 #define N 6
 87 typedef long long LL;
 88 typedef pair<int, int> Author;
 89 vector<pair<string, int> > VP; 
 90 
 91 int num[MAX] = {0}, p[MAX] = {0};
 92 int main(void){
 93     ios::sync_with_stdio(false); cin.tie(0);
 94 //  freopen("in.txt", "r", stdin);
 95 //  freopen("out.txt", "w", stdout);
 96     int i, j, a, b, ans;
 97     
 98     //素数筛选法 
 99     for(i = 2; i <= MAX; i++) if(!num[i]) for(j = i + i; j <= MAX; j += i) num[j] = 1; 
100     //p[0]存多少素数,p[i]存哪个素数 
101     for(i = 2; i <= MAX; i++) if(!num[i])p[++p[0]] = i;
102     
103     while(cin>>a>>b){
104         for(i = 1, ans = 0; i <= p[0]; i++){
105             int t = b / p[i];    //最大的能除的数 ,这里包含1, 2; 2, 4
106             if(t >= a)  ans += (t - a + 1);
107             else break;    //超过b; 
108         }
109         cout<<ans<<endl;
110     }
111 
112     return EXIT_SUCCESS;
113 }
View Code

 

 

链接:https://www.nowcoder.com/acm/contest/96/E
来源:牛客网

时间限制:C/C++ 1秒,其他语言2秒
空间限制:C/C++ 32768K,其他语言65536K
64bit IO Format: %lld

题目描述

  AA的欧尼酱qwb是个考古学家,有一天qwb发现了只白白圆圆小小的木乃伊,它是个爱哭鬼却很努力。qwb想把这么可爱的小木乃伊送给
AA,于是便找上了快递姐姐,这下可让快递姐姐犯愁了,因为去往AA家的路实在太难走了(甚至有可能没有路能走到AA家),快递姐姐
找上聪明的ACMer,想请你帮忙找出最快到达AA家的路,你行吗?

输入描述:

第一行输入两个整数n和m(2<=n<=m<=200000),分别表示有n座城市和m条路,城市编号为1~n(快递姐姐所在城市为1,AA所在城市为n)。
接下来m行,每行输入3个整数u,v,w(u,v<=n,w<=100000),分别表示城市u和城市v之间有一条长为w的路。

输出描述:

输出结果占一行,输出快递姐姐到达AA家最短需要走多远的路,如果没有路能走到AA家,则输出“qwb baka”(不用输出双引号)。
示例1

输入

4 4
1 2 1
2 3 2
3 4 3
2 3 1

输出

5

分析:看到神仙的DP解题,也有神仙用Spfa和Dijkstra过。

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 58                                        ;BBQQZaSJ5J15SJDQ7                                    .iEBL:.  . .,:,:;SRQGZXPXXSHaS2UUSUSswQB6OZgG6pZZGODEOGDE
 59                                       JgUri1aGEpEpXSS5LBQ,                              .rLap2Jr     ;712OO6ZggRDZXHXpKK2211LLLc7wgBa2PODRGE6OZgDgGDGD
 60                                     7QB; ,Jc76DaZXOZgDPEBBX,                       :c1HEZ1c;:,.,:;cHDgQQggMZgGOZO6OEGpXsLLsJXHPOBBBc;vss5EMggZDODOgED6
 61                                    LZJ::iwrrr72EPgXU5OBBBBBBBp7:               .i5P6wL;,   .:LS6DMgMgZKgEOPOGGDDGEXULvLSOBBBBBQBgDQDJ5rrr7JOQQDDZGEgDD
 62                                    wp rR5,       .HQRX7r72RQBBBBBQBMEK6Uc7sc7cHOU:,     .:LHgOQg6ZOPZpZGGEDERZPwcr7LKDDaJr:.      rBX;;:r:ir1DBQMgRgDp
 63                                    7D,sw:      ,   Z.      .LgBBQBBOsJaQBBBBBBXrr.  .:rwZBBQgROgDgDOERRMgROOScrLsPP5r,             7Qs,::::::rUgRgZOG6
 64                                    ,Q57:r  ,:  ,r  U:         .;;   .vL7L;r:UG7.:sr;JSgQBgDEZXXUSXpHa21svr7r7s2v:.                  iQK...:,:::;LLJJws
 65                                     OBHs;.  :;;,v  H:             :6X;       rpL;7pSrrr7r;::,....,,:;iirrvvvr,                       :BBPs;;:,,,,::;::
 66                                     ::rPDEL:..7:c;7r              pK  .:      :KL.:r:..     ..:icsS5sr:,.                              JgGgBQDOSJss77r
 67                                          :r;:::  ,                PH r,       ,;5: ,::::;7Ls7Lc7;:                                         ,:7JP17rJUs
 68                                                                   .:J: :;  .,:K6BQS7:.,.,.
 69                                                                     :r7Ji;r7;::;;.
 70                                                                         .                                                                             **/
 71                                                                         
 72 #include <bits/stdc++.h>
 73 using namespace std;
 74 
 75 #define FF(i, a, b) for(int i = a; i < b; i++)
 76 #define RR(i, a, b) for(int i = a; i > b; i++)
 77 #define ME(a, b) memset(a, b, sizeof(a))
 78 #define SC(x) scanf("%d", &x)
 79 #define PR(x) printf("%d\n", x)
 80 #define INF 0x3f3f3f3f
 81 #define MAX 200005
 82 #define MAXNUM 1e11
 83 #define MOD 1e9+7
 84 #define E 2.71828182845
 85 #define PI 3.1415926
 86 #define M 8
 87 #define N 6
 88 typedef long long LL;
 89 typedef pair<int, int> Author;
 90 vector<pair<string, int> > VP; 
 91 
 92 struct Node{
 93     LL u, v, w;
 94 };
 95 LL DP[MAX];
 96 
 97 int main(void){
 98     ios::sync_with_stdio(false); cin.tie(0);
 99     int i, n, m, flag, uu, vv, ww;
100     vector<Node> v;Node node;
101     ME(DP, 125); DP[1] = 0;flag = 1;
102     
103     cin>>n>>m;
104     for(i = 0; i < m; i++){
105         cin>>node.u>>node.v>>node.w;
106         v.push_back(node);
107     }
108     
109     while(flag){    //²»¶Ï¼ì²é£¬ÕÒµ½×îÓÅ·¾¶ 
110         flag = 0;
111         for(i = 0; i < v.size(); i++){
112             uu = v[i].u;vv = v[i].v; ww = v[i].w;  
113             //Ë«Ïò¼ì²é 
114             if(DP[uu] + ww < DP[vv]){
115                 DP[vv] = ww + DP[uu];
116                 flag = 1;
117             }
118             if(DP[vv] + ww < DP[uu]){
119                 DP[uu] = ww + DP[vv];
120                 flag = 1;
121             }
122         }
123     }
124     if(DP[n] > MAXNUM) cout<<"qwb baka"<<endl;
125     else cout<<DP[n]<<endl;
126 
127     return EXIT_SUCCESS;
128 }
View Code

链接:https://www.nowcoder.com/acm/contest/96/G
来源:牛客网

时间限制:C/C++ 1秒,其他语言2秒
空间限制:C/C++ 32768K,其他语言65536K
64bit IO Format: %lld

题目描述

给你一个n*m的图,地图上'.'代表可以走的地方,而'#'代表陷阱不能走,
'P'代表人物位置,'K'代表钥匙,'E'代表出口。人物一个,钥匙有多个,
('K'的数量<=50)),出口一个,每个位置可以向(上,下,左,右)四个
方向走一格,花费一个单位时间,现在你需要花费最少的时间拿到钥匙
然后从迷宫的出口出去(若没有钥匙,则不能进入迷宫出口所在的格子)。

输入描述:

第一行一个整数T(T <= 50),代表数据的组数
接下来一行n,m(n<=500,m<=500),代表地图的行和列
接下来n行,每行一个长度为m的字符串,组成一个图。

输出描述:

如果可以出去,输出所花费的最少时间。
如果不能出去,输出一行"No solution"。
示例1

输入

3
5 5
....P
##..E
K#...
##...
.....
5 5
P....
.....
..E..
.....
....K
5 5
P#..E
.#.#.
.#.#.
.#.#.
...#K

输出

No solution
12
No solution

分析:单向BFS+三维数组特判就可以了,也有人用双向BFS找最短路

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 58                                        ;BBQQZaSJ5J15SJDQ7                                    .iEBL:.  . .,:,:;SRQGZXPXXSHaS2UUSUSswQB6OZgG6pZZGODEOGDE
 59                                       JgUri1aGEpEpXSS5LBQ,                              .rLap2Jr     ;712OO6ZggRDZXHXpKK2211LLLc7wgBa2PODRGE6OZgDgGDGD
 60                                     7QB; ,Jc76DaZXOZgDPEBBX,                       :c1HEZ1c;:,.,:;cHDgQQggMZgGOZO6OEGpXsLLsJXHPOBBBc;vss5EMggZDODOgED6
 61                                    LZJ::iwrrr72EPgXU5OBBBBBBBp7:               .i5P6wL;,   .:LS6DMgMgZKgEOPOGGDDGEXULvLSOBBBBBQBgDQDJ5rrr7JOQQDDZGEgDD
 62                                    wp rR5,       .HQRX7r72RQBBBBBQBMEK6Uc7sc7cHOU:,     .:LHgOQg6ZOPZpZGGEDERZPwcr7LKDDaJr:.      rBX;;:r:ir1DBQMgRgDp
 63                                    7D,sw:      ,   Z.      .LgBBQBBOsJaQBBBBBBXrr.  .:rwZBBQgROgDgDOERRMgROOScrLsPP5r,             7Qs,::::::rUgRgZOG6
 64                                    ,Q57:r  ,:  ,r  U:         .;;   .vL7L;r:UG7.:sr;JSgQBgDEZXXUSXpHa21svr7r7s2v:.                  iQK...:,:::;LLJJws
 65                                     OBHs;.  :;;,v  H:             :6X;       rpL;7pSrrr7r;::,....,,:;iirrvvvr,                       :BBPs;;:,,,,::;::
 66                                     ::rPDEL:..7:c;7r              pK  .:      :KL.:r:..     ..:icsS5sr:,.                              JgGgBQDOSJss77r
 67                                          :r;:::  ,                PH r,       ,;5: ,::::;7Ls7Lc7;:                                         ,:7JP17rJUs
 68                                                                   .:J: :;  .,:K6BQS7:.,.,.
 69                                                                     :r7Ji;r7;::;;.
 70                                                                         .                                                                             **/
 71                                                                         
 72 #include <bits/stdc++.h>
 73 using namespace std;
 74 #define FF(i, a, b) for(int i = a; i < b; i++)
 75 #define RR(i, a, b) for(int i = a; i > b; i++)
 76 #define ME(a, b) memset(a, b, sizeof(a))
 77 #define SC(x) scanf("%d", &x)
 78 #define PR(x) printf("%d\n", x)
 79 #define INF 0x3f3f3f3f
 80 #define MAX 1001
 81 #define MOD 1e9+7
 82 #define E 2.71828182845
 83 #define PI 3.1415926
 84 #define M 8
 85 #define N 6
 86 typedef long long LL;
 87 typedef pair<int, int> Author;
 88 vector<pair<string, int> > VP;
 89 
 90 int n, k;
 91 int dis[MAX][MAX][2];    //ÓÐûÓÐÔ¿³× 
 92 int dir[4][2] = {1, 0, 0, 1, -1, 0, 0, -1};
 93 char matrix[MAX][MAX];
 94 typedef struct Node{
 95     int x, y, flag, step;    //step¼ÆËã·³Ì,flagÊÇÓÐÎÞÔ¿³× 
 96 }node;
 97 
 98 bool check(int x, int y){
 99     if(x >= 0 && x < n && y >= 0 && y < k) return 1;
100     return 0;
101 }
102 
103 int BFS(int x, int y){
104     queue<Node> p;Node temp, temp2;int i;
105     temp.x = x; temp.y = y; temp.step = 0; temp.flag = 0;
106     dis[x][y][0] = 1;    //³õʼλÖÃ
107     
108     p.push(temp);
109     while(!p.empty()){
110         temp2 = p.front(); p.pop();
111         if(matrix[temp2.x][temp2.y] == 'E') return temp2.step;    //Óöµ½ÖÕµã¾ÍÍ˳ö
112         for(i = 0; i < 4; i++){
113             int xx = temp2.x + dir[i][0];
114             int yy = temp2.y + dir[i][1];
115             
116             if(check(xx, yy) && matrix[xx][yy] != '#'){
117                 Node temp3;
118                 if(matrix[xx][yy] == 'K') temp3.flag = 1;    //ÕÒµ½Ô¿³×
119                 else temp3.flag = temp2.flag;
120                 if(dis[xx][yy][temp3.flag]) continue;    //Èô¸Ãµã·ÃÎʹý£¬ÔòÖ±½Ócontinue 
121                 dis[xx][yy][temp3.flag] =  1;
122                 if(matrix[xx][yy] =='E' && !dis[xx][yy][1]) continue;    //ÈôÕÒµ½ÖÕµãÇÒû±»·ÃÎʹý£¬Ôò½áÊø£¬·ÀÖ¹¾­¹ýÖÕµã 
123                 temp3.x = xx; temp3.y = yy; temp3.step = temp2.step + 1; p.push(temp3); 
124             }
125         }
126     }
127     return -1;
128 }
129 
130 
131 int main(void){
132     int i, j, T, bx, by;
133     
134     cin>>T;
135     while(T--){
136         ME(dis, 0);
137         cin>>n>>k;
138         for(i = 0; i < n; i++){
139             cin>>matrix[i];
140             for(j = 0; j < k; j++)if(matrix[i][j] == 'P'){bx = i; by = j;}
141         }
142         int ret = BFS(bx, by);
143         if(ret != -1) cout<<ret<<endl;
144         else  cout<<"No solution"<<endl;
145     }
146     
147     return EXIT_SUCCESS;
148 }
View Code

链接:https://www.nowcoder.com/acm/contest/96/J
来源:牛客网

时间限制:C/C++ 1秒,其他语言2秒
空间限制:C/C++ 32768K,其他语言65536K
64bit IO Format: %lld

题目描述

一天durong同学买了一个无限长的杯子,同时买了n个球,并且标号为1,2,3......n,durong同学突然想到一个问题----如果他把n个球依次,也就是按照1,2,3...n的顺序放进杯子里,然后在全部拿出来(注意不一定要等到全部放进去才能拿出球),并且会记录放进和拿出球的顺序,
durong想知道,要满足当第m个球进去后,杯子中此时恰好有k个球,然后仍然要把剩下的n-m个球放进去,最后杯中的球要取光,这样的放进和拿出球的顺序有多少种,答案有可能很大,所以mod上1e9+7

输入描述:

1<=n,m,k<=1e6(m可能大于n,k可能大于m)
第一行一个正整数T,表示数据组数。(1<=T<=10000)
对于每组数据包含一行三个正整数n,m,k。

输出描述:

对于每组数据输出一个正整数表示答案。
由于答案可能过大,所以只需要输出对1e9+7取模后的答案
示例1

输入

2
3 3 3
3 3 2

输出

1
2

分析:反向思维,第m个球不算,前面共m - 1个球,已放好k个球,需要放m - k 个球,第二步,放n - m个球,取n - m + k个球,卡特兰数公式判断即可,注意数组范围和逆元,不然会段错误

  1 /**
  2  
  3                                                                        :;LaEaHKEEGpPXU7;,
  4                                                                   .:75pKH11252U252XapZgRQgD6XJscLr;,.
  5                                                                :LXpRgGaX521JLw1JswJJsJs22XHPPEZEGDOMMRDOa7.
  6                                                            .r2EDDZEpZPZP6KpHX5SXH5XXa5KwaXaSX5UJ1c77sLs2GMQQ6r                                       .
  7                                                         ,LpgOGpEZGZEZEpZKpHHU5wP5HEDgpXpHa2SSa5aSXULr7rrirrJXRBp;                                   ;B
  8                                                      ,J6MRZH6EgEEZE6E6EZZPZXXwSSGQXr::aPpP5USUHaHaKa5Lvrr7ri;rLHBB2:                                Kc
  9                                                    rpQDOpPPOGGZOGOZG6GEOEOEDPPGBa.  .PaSSUXSUUUaUSaKXKS177r7rrrirSBBR7                             .O,
 10                                                 :UBQOKPK6ZOOOEDEO6GZE6EpEpDgDBR:   UBpXHa5aSaUS5SUS5XapPHJc7rrv7rr7sgBBs                           .g.
 11                                               ;gBMPXpO6GEOEOEOEGEOEE6EZEEDRGBB    EB5pKSXpHKaHSX552S5aUHHEX17c7vr7777s5RBS:                        .R;
 12                                            .sQBPXpDZOODOgODGOGgEDEOEOGgGgOOMB:   LBKKSXSHa52aaKXHXKaa5aSaaHXSJLcL7vcc777JDBBg2;.                    Qi
 13                                          ;2ggp2EDDOggGEDGgDDOgGGZDOOZOGg6gEBX    vBZaHUKaaUXXXSXXKXpXHXH5wwaa52U1wssLsLJccv1UDQBQ67.                O7
 14                                        :ZZUU5PROOEOZOZGGOODZOGgODZOOgOggRgRB; ;:..R6XaKKpP6PGppKPHpHpPX5aU21UUa5Sw52UwUJJv77L77sSpQMDU;            ;B5
 15                                      ,SRJ7sSHGggEOZOEG6OODpOZgggOQQBBBBBQBBQ.,;;. LBOgOgDRDDZODMgQRgRDaa552a252UUa25w5UaU2sLvccs7r7sJZBBMr       ,XQJ:
 16                                     LQHr77J6RGOZDZOEGEDGgDRORQBBBQRDPU1Jscwa7.,::.:J7r;::::.  ..,:;i7UOgRRRgDPH5SUSU52U2HHa1JJJLJLccLr71RBB,    7R2,
 17                                   :RZv;77JSgGOZEODEDGOEggQBBBMS7;:,:,,.,.,.:7L;:,;.:  ,: . .....       .:;rJU6GgGRggEZHPaKXX2S221Js1Lc7r:7QB. .XX:
 18                                  7g1;;7rcXG6gpGDZGgZOOQBBQpr.   ::::;:;:::Jr::sr;::;:.:vs,:::::::::::,,         ,:7L5HGOggRgZUUU5wSUaJLc7r7BOiDr  ...
 19                                .XX;;;irLHGKpZZZEKgDRBBB6i    ,;;:;;;;r;;:s177:,;L7:;7:.rHi,:::::::::::::::::,.         .,;7ZRQgO6KUUJsLwsJ7KBM. .....
 20                                JZ:;rrrc5EPHp6XgpRBBBE;     :i:;;;;;:;;;:c7::r7;,::::::::rv:,:::::::::::,:,:,:,,,,            .;sORQRGX21wsXU:  .... .
 21                               .Br;iir72EHPHZ6EgBBR7.   ..;ii:;;;;;;;;;:71r::.:7,  .::7;:;H;.::,:j::::::,:::::,,,::,,,              ,7wEDRZBMr  .......
 22                               1D;:r;rwOPXPKH6BBX,   .:;:;;;:::;:;;;;;::Ls,;ss..r.  ,c77;sLU,:,:::,:::,:,:,:,,,:,,,:::.                  .:.rP:.......
 23                               D2:i;rJpKHXHXgg7    .,,::::;:::::;:;:;;:;SL7sS2, :.   ::::,:U7.:::,:::,:,,,,,:,,.,,,.,:,                    ;L:. .. ...
 24                              :Qc;i7LGZPa6gBM,  ...,,::::::::::::;;::;.JJ;ic:         ;:::,v1,.:::;7,,,:,:,:,,,,.:,,,,..                :2wr.  .......
 25                              sRr:rrwZGgBQR7.  .,,::::::::::::::;::;:::Hr:7i          ,;;:::U: .,,,:r.,,:,:,:,,,:,,,.....             7K2:. ..........
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 41                            J7  rEBBg..,. ..... .  ,L1PLKr:,:;r;;::, :,                       :;SXsJU1XLLr....:,,,,,,::::,::::::,::LL,,rZXUJU25552aUUUU
 42                            JL .:sXBB:... ., .    ,OH777;,rZRBBBQL ..r,                     ,:r7::r:,;;....:,,,,,,,,.::::;::;:::.7gL    UK2UUS1aw2wU25J
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 46                            5; ;;r2QBB,  .s:  :a7:PBZ2,:BEaZPKgBZOgB,                        .   ...:LEHri::,:.,,.,,,::rsXRBgGKEJi,    2GH252SUS5S11wH5
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 51                           1i :SZJrLXpBBMRB:  .v  Ba,.       :sv:                             .        irsBUSpEEGPPpg65r:..,;sKZDZHSX5XaHU55K2wUGMRZ6Z6
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 56                            ,cs5aXaP552ssLwRSHXaSS5asXBB.      . .                             .  LBBgGgZHsr;vrr;rcXpP5UUS5X5S2XS5SS1URBggpEZgDDGOGggRD
 57                               ,;irrs2KgQRPJJJSUXKXSHJpBM   . . .                                ZBBOUsr;:.  ,,,rHgZPUaSaUSUSSXaUPXsSBB6ZZp6EEGOOGDEggO
 58                                        ;BBQQZaSJ5J15SJDQ7                                    .iEBL:.  . .,:,:;SRQGZXPXXSHaS2UUSUSswQB6OZgG6pZZGODEOGDE
 59                                       JgUri1aGEpEpXSS5LBQ,                              .rLap2Jr     ;712OO6ZggRDZXHXpKK2211LLLc7wgBa2PODRGE6OZgDgGDGD
 60                                     7QB; ,Jc76DaZXOZgDPEBBX,                       :c1HEZ1c;:,.,:;cHDgQQggMZgGOZO6OEGpXsLLsJXHPOBBBc;vss5EMggZDODOgED6
 61                                    LZJ::iwrrr72EPgXU5OBBBBBBBp7:               .i5P6wL;,   .:LS6DMgMgZKgEOPOGGDDGEXULvLSOBBBBBQBgDQDJ5rrr7JOQQDDZGEgDD
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 63                                    7D,sw:      ,   Z.      .LgBBQBBOsJaQBBBBBBXrr.  .:rwZBBQgROgDgDOERRMgROOScrLsPP5r,             7Qs,::::::rUgRgZOG6
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 65                                     OBHs;.  :;;,v  H:             :6X;       rpL;7pSrrr7r;::,....,,:;iirrvvvr,                       :BBPs;;:,,,,::;::
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 68                                                                   .:J: :;  .,:K6BQS7:.,.,.
 69                                                                     :r7Ji;r7;::;;.
 70                                                                         .                                                                             **/
 71                                                                         
 72 #include <bits/stdc++.h>
 73 using namespace std;
 74 
 75 #define FF(i, a, b) for(int i = a; i < b; i++)
 76 #define RR(i, a, b) for(int i = a; i > b; i++)
 77 #define ME(a, b) memset(a, b, sizeof(a))
 78 #define SC(x) scanf("%d", &x)
 79 #define PR(x) printf("%d\n", x)
 80 #define INF 0x3f3f3f3f
 81 #define MAX 4100000
 82 #define MOD 1000000007
 83 #define E 2.71828182845
 84 #define PI 3.14159265358979
 85 #define M 8
 86 #define N 6
 87 typedef long long LL;
 88 typedef pair<int, int> Author;
 89 vector<pair<string, int> > VP; 
 90 
 91 LL fac[MAX];
 92 //标准快速幂 
 93 LL qkm(LL a, LL b, LL mod){
 94     LL ans = 1;
 95     for(a %= mod; b; b >>= 1, a = a * a % mod)if(b & 1)ans=ans*a%mod;
 96     return ans;
 97 }
 98 
 99 LL C(LL a, LL b){
100     return fac[a] * qkm(fac[b], MOD - 2, MOD) % MOD * qkm(fac[a - b], MOD - 2, MOD) % MOD;
101 }
102 
103 //卡特兰数 
104 LL Catalan(LL n, LL m){
105     if(m == 0) return 1;
106     else return (C(n + m, n) - C(n + m, n + 1) + MOD) % MOD;        //被CF坑的 
107 }
108 
109 int main(void){
110     ios::sync_with_stdio(false); cin.tie(0);
111     LL i, T, n, m, k;
112     cin>>T;
113     
114     for(fac[0] = 1, fac[1] = 1, i = 2; i <= MAX; i++)fac[i] = (fac[i - 1] * i) % MOD;
115     while(T--){
116         cin>>n>>m>>k;
117         if(m > n || k > m) cout<<0<<endl; 
118         else cout<<(Catalan(m - 1, m - k) * Catalan(n - m + k, n - m))% MOD<<endl;    //逆向思维 
119     }
120     
121     return EXIT_SUCCESS;
122 }
View Code

链接:https://www.nowcoder.com/acm/contest/96/L
来源:牛客网

题目描述

集训队暑训的时候,是睡在机房的,有几个小伙子不甘寂寞,带了仓鼠来机房。有的还不止带了一只!
为此,贴心的集训队刘队长买了一大一小两种放仓鼠笼子的架子,大的可以放两个仓鼠笼 小的只能放一个。
要注意的是仓鼠是领地动物,一般来讲 仓鼠笼与仓鼠笼之间是不能贴在一起的,
但同一个人带来的仓鼠的仓鼠笼可以被放在一起(既两个仓鼠笼如果属于同一个人,那么这两个仓鼠笼就可以被放在同一个大架子上)
现在告诉你现有的两种架子的数量和多少人带了仓鼠以及每个人带的仓鼠的数量,问你是否能将这些仓鼠放下。

输入描述:

第一行为一个数T 表示测试样例组数

对于每组测试数据

第1行输入a,b,n; 分别为小架子的数量,大架子的数量,带了仓鼠的人数

第2行有n个数字P
1
-P
n
,分别表示每个人带的仓鼠的数量

输出描述:

如果放得下 则输出"Yes" 否则输出"No"
示例1

输入

1
1 2 4
1 1 1 1

输出

No

分析:数学思维,先对所有人计算最多大架子数,再判断比实际大架子数量多还是少,多就需要双倍小架子,少就需要把剩余大架子数当成小架子计算

  1 /**
  2  
  3                                                                        :;LaEaHKEEGpPXU7;,
  4                                                                   .:75pKH11252U252XapZgRQgD6XJscLr;,.
  5                                                                :LXpRgGaX521JLw1JswJJsJs22XHPPEZEGDOMMRDOa7.
  6                                                            .r2EDDZEpZPZP6KpHX5SXH5XXa5KwaXaSX5UJ1c77sLs2GMQQ6r                                       .
  7                                                         ,LpgOGpEZGZEZEpZKpHHU5wP5HEDgpXpHa2SSa5aSXULr7rrirrJXRBp;                                   ;B
  8                                                      ,J6MRZH6EgEEZE6E6EZZPZXXwSSGQXr::aPpP5USUHaHaKa5Lvrr7ri;rLHBB2:                                Kc
  9                                                    rpQDOpPPOGGZOGOZG6GEOEOEDPPGBa.  .PaSSUXSUUUaUSaKXKS177r7rrrirSBBR7                             .O,
 10                                                 :UBQOKPK6ZOOOEDEO6GZE6EpEpDgDBR:   UBpXHa5aSaUS5SUS5XapPHJc7rrv7rr7sgBBs                           .g.
 11                                               ;gBMPXpO6GEOEOEOEGEOEE6EZEEDRGBB    EB5pKSXpHKaHSX552S5aUHHEX17c7vr7777s5RBS:                        .R;
 12                                            .sQBPXpDZOODOgODGOGgEDEOEOGgGgOOMB:   LBKKSXSHa52aaKXHXKaa5aSaaHXSJLcL7vcc777JDBBg2;.                    Qi
 13                                          ;2ggp2EDDOggGEDGgDDOgGGZDOOZOGg6gEBX    vBZaHUKaaUXXXSXXKXpXHXH5wwaa52U1wssLsLJccv1UDQBQ67.                O7
 14                                        :ZZUU5PROOEOZOZGGOODZOGgODZOOgOggRgRB; ;:..R6XaKKpP6PGppKPHpHpPX5aU21UUa5Sw52UwUJJv77L77sSpQMDU;            ;B5
 15                                      ,SRJ7sSHGggEOZOEG6OODpOZgggOQQBBBBBQBBQ.,;;. LBOgOgDRDDZODMgQRgRDaa552a252UUa25w5UaU2sLvccs7r7sJZBBMr       ,XQJ:
 16                                     LQHr77J6RGOZDZOEGEDGgDRORQBBBQRDPU1Jscwa7.,::.:J7r;::::.  ..,:;i7UOgRRRgDPH5SUSU52U2HHa1JJJLJLccLr71RBB,    7R2,
 17                                   :RZv;77JSgGOZEODEDGOEggQBBBMS7;:,:,,.,.,.:7L;:,;.:  ,: . .....       .:;rJU6GgGRggEZHPaKXX2S221Js1Lc7r:7QB. .XX:
 18                                  7g1;;7rcXG6gpGDZGgZOOQBBQpr.   ::::;:;:::Jr::sr;::;:.:vs,:::::::::::,,         ,:7L5HGOggRgZUUU5wSUaJLc7r7BOiDr  ...
 19                                .XX;;;irLHGKpZZZEKgDRBBB6i    ,;;:;;;;r;;:s177:,;L7:;7:.rHi,:::::::::::::::::,.         .,;7ZRQgO6KUUJsLwsJ7KBM. .....
 20                                JZ:;rrrc5EPHp6XgpRBBBE;     :i:;;;;;:;;;:c7::r7;,::::::::rv:,:::::::::::,:,:,:,,,,            .;sORQRGX21wsXU:  .... .
 21                               .Br;iir72EHPHZ6EgBBR7.   ..;ii:;;;;;;;;;:71r::.:7,  .::7;:;H;.::,:j::::::,:::::,,,::,,,              ,7wEDRZBMr  .......
 22                               1D;:r;rwOPXPKH6BBX,   .:;:;;;:::;:;;;;;::Ls,;ss..r.  ,c77;sLU,:,:::,:::,:,:,:,,,:,,,:::.                  .:.rP:.......
 23                               D2:i;rJpKHXHXgg7    .,,::::;:::::;:;:;;:;SL7sS2, :.   ::::,:U7.:::,:::,:,,,,,:,,.,,,.,:,                    ;L:. .. ...
 24                              :Qc;i7LGZPa6gBM,  ...,,::::::::::::;;::;.JJ;ic:         ;:::,v1,.:::;7,,,:,:,:,,,,.:,,,,..                :2wr.  .......
 25                              sRr:rrwZGgBQR7.  .,,::::::::::::::;::;:::Hr:7i          ,;;:::U: .,,,:r.,,:,:,:,,,:,,,.....             7K2:. ..........
 26                              OX:irsXgQZ:.   .,,:::,,:::,::::::::;;;::r5;r7:           :;;;:7L ..,.,;:,:,:,,,:,:,,,,......        .rU6w;..............
 27                             .BJ71EK5;.   .,,::,:,:,,,:,:,::::::::;:;.Ls;r7             :;:::s, ..,,,r:.:.,,,.:, .,..... .    .;s5XJr,..,.............
 28                             1Mv::.     .:,:,,,,,,,:::::::,;:::;:;::..J7;c:          ,. ,rri:27  , .,:;. ,:,:,,:..,....    .rpPL;:.. ... .............
 29                        ..7Ls:        .,,,:,,,:,:,:::,:,:::::,:::::: .Srrr,  .,:;;;::::..:7r;r1    .: :E:..,,:,...,,.., ..XBQ7, ................... ..
 30                   ,;7v7r7:,         ,,,,:.,,,,,,:,,::,:,,.::.,:.,:, :Jrr; ,r7,:..        ,:::L:  .  .7RJ .,.:.,......,,:MBs   ..............,........
 31             .:;vJs7i,              ,,,., ..,,:,,,:,,,:,:.,rs,,.:,,  ;J;c,                .,::;;     Lr.E: .,...,....:,:1Z:  .........,.,...........:,.
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 69                                                                     :r7Ji;r7;::;;.
 70                                                                         .                                                                             **/
 71                                                                         
 72 #include <bits/stdc++.h>
 73 using namespace std;
 74 
 75 #define FF(i, a, b) for(int i = a; i < b; i++)
 76 #define RR(i, a, b) for(int i = a; i > b; i++)
 77 #define ME(a, b) memset(a, b, sizeof(a))
 78 #define SC(x) scanf("%d", &x)
 79 #define PR(x) printf("%d\n", x)
 80 #define INF 0x3f3f3f3f
 81 #define MAX 1001
 82 #define MOD 1e9+7
 83 #define E 2.71828182845
 84 #define PI 3.1415926
 85 #define M 8
 86 #define N 6
 87 typedef long long LL;
 88 typedef pair<int, int> Author;
 89 vector<pair<string, int> > VP; 
 90 
 91 int main(void){
 92     ios::sync_with_stdio(false); cin.tie(0);
 93     LL T, odd, even, i, a, b, n, num, flag;
 94     
 95     cin>>T;
 96     while(T--){
 97         cin>>a>>b>>n; odd = even = 0;flag = 0;
 98         for(i = 0; i < n; i++){
 99             cin>>num;
100             even += (num / 2);
101             num %= 2;
102             odd += num;
103         }
104         if(b >= even) b -= even;
105         else{
106             flag = -1;
107             even -= b;
108         }
109         if(flag == 0){
110             if(a + b >= odd) cout<<"Yes"<<endl;
111             else cout<<"No"<<endl;
112         }
113         else {
114             if(a >= odd + even * 2) cout<<"Yes"<<endl;
115             else cout<<"No"<<endl;
116         }
117     }
118     
119     return EXIT_SUCCESS;
120 }
View Code

 





 

posted @ 2018-04-20 09:17  Empirefree  阅读(1432)  评论(0编辑  收藏  举报