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江西财经大学第一届程序设计竞赛题解

(！！！第一次抽奖中衣服，这个题解是必须要写的！！！)

A,D签到题就不讲了，C题注意题目中2人距离不变，所以时间t = L / v1即可

链接：https://www.nowcoder.com/acm/contest/115/B
来源：牛客网

大吉大利
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

给出一个出生日期，比如：1999-09-09，
问：从出生那一天开始起，到今天2018-04-21为止（包括出生日期和今天），有多少天，年月日都不包含数字4？

输入描述:

第一行输入一个整数T（表示样例个数）
接下来T组样例
每个样例一行，包含一个字符串“yyyy-mm-dd”（1990<=yyyy<=2018）
题目保证测试数据的正确性

输出描述:

输出题意要求的天数
示例1

输入

1
1999-09-09

输出

5020

分析：一开始是想直接每年每年算，结果直接崩了，首先看清题意，year从1990开始，所以year % 10 != 4即可,再循环到2018-4-21号，控制月份和年份以及相关条件即可







  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;
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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 1001
 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 
 92 int m[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};        //月份 
 93 bool leap(int y){
 94     return y % 4 == 0 && y % 100 != 0 || y % 400 == 0; 
 95 }
 96 
 97 int main(void){
 98     #ifdef LOCAL
 99         freopen("in.txt", "r", stdin);
100         freopen("out.txt", "w", stdout);
101     #endif
102     ios::sync_with_stdio(false); cin.tie(0);
103     int T, year, month, day, cnt;
104     
105     scanf("%d", &T);
106     while(T--){
107         scanf("%d-%d-%d", &year, &month, &day);cnt = 0;
108         if(leap(year))m[2] = 29;    //闰年
109         else m[2] = 28;
110         if(year % 10 != 4 && month != 4 && day % 10 != 4) cnt++;
111         
112         while(year != 2018 || month != 4 || day != 21){
113             day++;
114             //控制年，月，日进位
115             if(day > m[month]){day = 1;month++;}    
116             if(month > 12){month = 1; year++;if(leap(year)) m[2] = 29;else m[2] = 28;}
117             if(year % 10 != 4 && month != 4 && day % 10 != 4) cnt++;
118         }
119         printf("%d\n", cnt);
120     }
121     
122     return EXIT_SUCCESS;
123 }



View Code


链接：https://www.nowcoder.com/acm/contest/115/E
来源：牛客网





消息列表












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







当你的好友给你发来一条消息，你的消息列表上就会置顶显示该好友的名字以及该好友发给你的消息总数，换句话说，你的消息列表里的好友是按跟你发消息的时间进行排序的，给你发消息的时间离当前时间越近的好友将排到越前面。当然，你可能会手动置顶一些好友，那么其他的好友给你发消息时，他们的名字就只能在你手动置顶好友的后面再置顶了。如果消息被你查看或者忽略，又或者你把好友消息删除了，消息总数将重置为0。
 
 根据用户的需求，有以下几个功能，需要你来实现：
 （1）recv：收到一条好友消息，对于手动置顶好友的消息，将在“手动置顶好友列表”里置顶；对于其他好友的消息，将在“手动置顶好友列表”之下的消息列表里置顶，同时都需要显示该好友的消息总数。
 （2）view：查看好友消息，将使该好友消息数变为0。
 （3）up：手动置顶好友。
 （4）down：取消手动置顶。
 （5）delete：删除好友消息，这个操作将使该好友从消息列表中删除，同时取消对该好友的手动置顶（如果存在的话）。
 
 假设初始消息列表为空，经过了一系列好友消息的操作之后，最终的消息列表将是怎么样的呢？


输入描述:


第一行输入一个整数T（表示样例个数）
接下来T组样例。
每组样例
第一行输入一个整数M，表示操作数（1≤M≤1000000）；
接下来M行，
每行输入一个操作，由一个操作类型和一个好友id构成，之间以空格分开，操作类型如上面5个英文单词表示，
例如：“recv 123456”表示接收到id为123456的好友的一条消息，“delete 123456”表示在消息列表中删除 id 为123456的好友的消息记录。
为了简化问题，一开始消息列表为空并假设好友名字id由六位数字“唯一”标识（000000≤id≤999999），
题目保证输入数据的一致性。


输出描述:


每组样例，
输出最后的消息列表，自顶向下，每行输出一个：“好友id 消息数”。
每组样例后空一行。





示例1






输入




1
13
recv 000001
recv 000002
up 000002
view 000001
recv 000002
recv 000004
up 000004
up 000001
recv 000004
recv 000003
view 000001
view 000004
down 000002








输出




000004 0
000001 0
000003 1
000002 2



分析:说实话，第一次遇到这个sort类型题目，感觉挺实用的，和QQ消息差不多，加time,flag保证题目中“越接近现在排序越靠前”和“置顶比不置顶牛皮”即可，再就是最后格式输出，puts("");是AC，cout<<endl；会WA,另外ME(mes, 0)可以初始化结构体且内部变量可以直接++,另外别忽略id为000000即可






  1 /**
  2  
  3                                                                        :;LaEaHKEEGpPXU7;,
  4                                                                   .:75pKH11252U252XapZgRQgD6XJscLr;,.
  5                                                                :LXpRgGaX521JLw1JswJJsJs22XHPPEZEGDOMMRDOa7.
  6                                                            .r2EDDZEpZPZP6KpHX5SXH5XXa5KwaXaSX5UJ1c77sLs2GMQQ6r                                       .
  7                                                         ,LpgOGpEZGZEZEpZKpHHU5wP5HEDgpXpHa2SSa5aSXULr7rrirrJXRBp;                                   ;B
  8                                                      ,J6MRZH6EgEEZE6E6EZZPZXXwSSGQXr::aPpP5USUHaHaKa5Lvrr7ri;rLHBB2:                                Kc
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 10                                                 :UBQOKPK6ZOOOEDEO6GZE6EpEpDgDBR:   UBpXHa5aSaUS5SUS5XapPHJc7rrv7rr7sgBBs                           .g.
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 12                                            .sQBPXpDZOODOgODGOGgEDEOEOGgGgOOMB:   LBKKSXSHa52aaKXHXKaa5aSaaHXSJLcL7vcc777JDBBg2;.                    Qi
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 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. .....
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 59                                       JgUri1aGEpEpXSS5LBQ,                              .rLap2Jr     ;712OO6ZggRDZXHXpKK2211LLLc7wgBa2PODRGE6OZgDgGDGD
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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 1100000
 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 //Ò»ÌõÐÅÏ¢°´ÕÕÊÇ·ñÖö¥À´sort£¬Öö¥ºó°´ÕÕ½ÓÊÕÐÅÏ¢ÏȺóÀ´sort 
 92 struct Node{
 93     int id;
 94     int num;
 95     int flag;
 96     int time;
 97 }mes[MAX];
 98 
 99 
100 bool cmp(const Node& a, const Node& b){
101     if(a.flag != b.flag) return a.flag > b.flag;
102     else return a.time > b.time;
103 }
104 
105 int main(void){
106 //    #ifdef LOCAL
107 //        freopen("in.txt", "r", stdin);
108 //        freopen("out.txt", "w", stdout);
109 //    #endif
110     ios::sync_with_stdio(false); cin.tie(0);
111     int i, T, m, myid, cnt; string str;
112     
113     cin>>T;
114     while(T--){
115         cin>>m; cnt = 1;ME(mes, 0);
116         while(m--){
117             cin>>str>>myid;
118             mes[myid].id = myid;
119             if(str == "recv"){
120                 mes[myid].num++;
121                 mes[myid].time = cnt++;        //¼ÆËã½ÓÊÕʱ¼ä 
122             }
123             else if(str == "up")mes[myid].flag = 1;
124             else if(str == "view") mes[myid].num = 0;
125             else if(str == "down") mes[myid].flag = 0;
126             else if(str == "delete"){mes[myid].flag = mes[myid].num = mes[myid].time = 0;}
127         }
128         sort(mes, mes + MAX, cmp);
129         for(i = 0; i <= MAX; i++)if(mes[i].time)printf("%06d %d\n", mes[i].id, mes[i].num);
130         puts("");
131     }
132     
133     return EXIT_SUCCESS;
134 }



View Code


链接：https://www.nowcoder.com/acm/contest/115/F
来源：牛客网





解方程












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







对于方程 2018 * x ^ 4 + 21 * x + 5 * x ^ 3 + 5 * x ^ 2 + 14 = Y，
 告诉你Y的值，你能找出方程在0~100之间的解吗？


输入描述:


第一行输入一个正整数T（表示样例个数）
接下来T组样例
每组样例一行，输入一个实数Y


输出描述:


一行输出一个样例对应的结果，
输出方程在0~100之间的解，保留小数点后4位小数；如果不存在，输出 -1





示例1






输入




2
1
20180421








输出




-1
9.9993


分析:这题牛皮了，一开始想的是循环，可是发现0~100内可以为小数，二分还是牛逼。


 






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 57                               ,;irrs2KgQRPJJJSUXKXSHJpBM   . . .                                ZBBOUsr;:.  ,,,rHgZPUaSaUSUSSXaUPXsSBB6ZZp6EEGOOGDEggO
 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
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 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 1001
 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 double formula(double x){
 92     return 2018 * pow(x, 4) + 21 * x + 5 * pow(x, 3) + 5 * pow(x, 2) + 14;
 93 }
 94 
 95 
 96 int main(void){
 97     ios::sync_with_stdio(false); cin.tie(0);
 98 //  freopen("in.txt", "r", stdin);
 99 //  freopen("out.txt", "w", stdout);
100     int i, j, T;double y, left, mid, right;
101     
102     cin>>T;
103     while(T--){
104         cin>>y;    
105         left = formula(0) - y;right = formula(100) - y;
106         if(right * left > 0){cout<<-1<<endl;continue;}    //µ¥µ÷µÝÔöº¯Êý     
107         left = 0, right = 100;
108         while(right - left > 1e-7){
109             mid = (right + left) / 2;
110             if(formula(mid) - y < 0) left = mid;
111             else right = mid;
112         }
113         cout<<fixed<<setprecision(4)<<mid<<endl;
114     }
115 
116     return EXIT_SUCCESS;
117 }



View Code


链接：https://www.nowcoder.com/acm/contest/115/G
来源：牛客网





小Q的口袋校园












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









  周末，小Q喜欢在PU口袋校园上参加各种活动刷绩点，体验丰富多彩的大学生活。 


  但是每个活动有各自的开始时间、结束时间、Happy值以及绩点数，活动之间可能存在时间冲突。 


  小Q是一个认真踏实的女孩，她在同一时间只会参加一个活动。 


  给出每一天的活动数，求小Q在同一天内所能获得的最大Happy值 与 绩点数。 


  小Q是一个活泼开朗的女孩，她总是寻求最大的Happy值，如果Happy值相同，才会尽可能使绩点数更多。 





输入描述:


第一行输入一个整数T（表示样例个数）
接下来T组样例
每组样例第一行输入一个整数N（表示有几项活动）
接下来N行，每行输入s,e,h,p 4个整数，分别代表一个活动的开始时间、结束时间、Happy值以及绩点数
0<=s<e<=24,1<=h,p<=100，1<=T,N<=20


输出描述:


输出T行
一行输出一个样例对应的结果
小Q在一天内所能获得的最大 Happy值 与 绩点数





示例1






输入




1
4
1 3 1 1
2 5 2 1
3 7 2 1
5 8 1 2








输出




3 3








说明




有两种方案
1）选择 第1、3两个活动，可以获得 Happy值 3 和 绩点数 2；
2）选择 第2、4两个活动，可以获得 Happy值 3 和 绩点数 3.


分析：这个和活动安排问题有点像，算是加强版了吧，转移方程我是没推出来，DFS倒是骚了一波，预处理一下，排个序就可以了








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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 30
 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 struct Node{
 92     int b,e,h,p;
 93 }acvity[MAX];
 94 int happy, score, n;
 95 
 96 void DFS(int ha, int so, int num, int ee){
 97     if(happy < ha || (happy == ha && score < so)){happy = ha; score = so;}
 98     
 99     if(num >= n) return ;
100     if(acvity[num].b >= ee) DFS(ha + acvity[num].h, so + acvity[num].p, num + 1, acvity[num].e);
101     DFS(ha, so, num + 1, ee);
102 }
103 bool cmp(const Node& a, const Node& b){return a.b < b.b;}
104 
105 int main(void){
106     #ifdef LOCAL
107         freopen("in.txt", "r", stdin);
108         freopen("out.txt", "w", stdout);
109     #endif
110     ios::sync_with_stdio(false); cin.tie(0);
111     int i, j, T;
112     
113     cin>>T;
114     while(T--){
115         cin>>n;happy = score = 0;
116         for(i = 0; i < n; i++)cin>>acvity[i].b>>acvity[i].e>>acvity[i].h>>acvity[i].p;
117         sort(acvity, acvity + n, cmp);
118         DFS(0, 0, 0, 0);
119         cout<<happy<<" "<<score<<endl;
120     }
121 
122     return EXIT_SUCCESS;
123 }



View Code


 


链接：https://www.nowcoder.com/acm/contest/115/H
来源：牛客网





小P的数学问题








 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









  晚上，小P喜欢在寝室里一个个静静的学习或者思考，享受自由自在的单身生活。 


  他总是能从所学的知识散发出奇妙的思维。 


  今天他想到了一个简单的阶乘问题， 


  0！= 1 


  1！= 1 


  2！= 1 * 2 = 2 


  3！= 1 * 2 * 3 = 6 


  4！= 1 * 2 * 3 *4 = 24 


  5！= 1 * 2 * 3 *4 * 5 = 120 


  。 


  。 


  如果 n=1000000000，那么n的阶乘会是多少呢，小P当然知道啦，那么你知道吗？ 





输入描述:


第一行输入一个整数T（表示样例个数）
接下来T组样例
每组样例一行，输入一个整数N（0<=N<=1000000000）


输出描述:


输出T行
每一行输出N的阶乘 N！（由于这个数比较大，所以只要输出其对1000000007取膜的结果即可）





示例1






输入




2
0
1000000000








输出




1
698611116



分析：都对n!情有独钟啊，斯特灵公式求n!位数,或者数组模拟求n!（不%），或者大数n!，fft算法或者python,java大法,再就是这个题目了吧


，求n!要%MOD，板子题，没什么讲的


 


















 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 #define MAX 10000000
 4 #define MOD 1000000007
 5 #define LL long long
 6 
 7 LL ans[120]={1,682498929,491101308,76479948,723816384,67347853,27368307,625544428,199888908,
 8     888050723,927880474,281863274,661224977,623534362,970055531,261384175,195888993,66404266,
 9     547665832,109838563,933245637,724691727,368925948,268838846,136026497,112390913,135498044,
10     217544623,419363534,500780548,668123525,128487469,30977140,522049725,309058615,386027524,
11     189239124,148528617,940567523,917084264,429277690,996164327,358655417,568392357,780072518,
12     462639908,275105629,909210595,99199382,703397904,733333339,97830135,608823837,256141983,
13     141827977,696628828,637939935,811575797,848924691,131772368,724464507,272814771,326159309,
14     456152084,903466878,92255682,769795511,373745190,606241871,825871994,957939114,435887178,
15     852304035,663307737,375297772,217598709,624148346,671734977,624500515,748510389,203191898,
16     423951674,629786193,672850561,814362881,823845496,116667533,256473217,627655552,245795606,
17     586445753,172114298,193781724,778983779,83868974,315103615,
18     965785236,492741665,377329025,847549272,698611116};
19 
20 
21 
22 int main(void){
23     LL i, T, n, m, cnt;
24     
25     cin>>T;
26     while(T--){
27         cin>>n;
28         m = n / MAX;
29         cnt = ans[m];
30         
31         for(i = m * MAX + 1; i <= n; i++)cnt = (cnt * i) % MOD;
32         cout<<cnt<<endl;
33     }
34 
35     return EXIT_SUCCESS;
36 }



View Code


链接：https://www.nowcoder.com/acm/contest/115/I
来源：牛客网





小P和小Q












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









  小P和小Q是好朋友，今天他们一起玩一个有趣的游戏。 


  他们的初始积分都为1，赢的人可以将自己的分数乘以 (K的平方)，而输的人也能乘以K。 


  他们玩的太开心了，以至于忘了自己玩了多久，甚至 K 是多少和游戏进行的回合数 N 都忘了。 


  现在给出他们俩最终的积分a,b，请问是否存在正整数K、N满足这样的积分，判断他们的游戏结果是否可信。 





输入描述:


第一行输入一个整数T（表示样例个数）
接下来T组样例
每组样例一行，输入两个正整数a,b（0<a,b<=1e9）


输出描述:


输出T行
一行输出一个样例对应的结果
若结果可信，输出 Yes
否则，输出 No





示例1






输入




6
2 4
75 45
8 8
16 16
247 994
1000000000 1000000








输出




Yes
Yes
Yes
No
No
Yes










备注:


每回合的K可能不同







分析：这题我感觉是真的有意思，小p：k^(2a) * k^(n - a)；小q:k^(2n - 2a) * k^(a);所以二者相乘k ^ 3n，说明k,n一定为整数即(int)(cbrt(a * b) + 0.5)若不等于(a * b)^3者说明不存在k,n








 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 #define MAX 10000000
 4 #define MOD 1000000007
 5 #define LL long long
 6 
 7 int main(void){
 8     LL i, a, b, T, ans;
 9     
10     //由题意推得k,n应为正数 
11     cin>>T;
12     while(T--){
13         cin>>a>>b;
14         ans = (LL)(cbrt(a * b) + 0.5);
15         if(ans * ans * ans == a * b)    cout<<"Yes"<<endl;
16         else cout<<"No"<<endl;
17     }
18     
19     return EXIT_SUCCESS;
20 }



View Code


 














 















































            

            
posted @ 
2018-05-05 12:00 
Empirefree 
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