题意：给出n，求出两个合数x和y使x+y=n。

题解：暴力筛合数，然后暴力找

 1 //#pragma comment(linker, "/STACK:102400000,102400000")
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<iostream>
 5 #include<cstring>
 6 #include<algorithm>
 7 #include<cmath>
 8 #include<map>
 9 #include<set>
10 #include<stack>
11 #include<queue>
12 using namespace std;
13 #define ll __int64
14 #define usint unsigned int
15 #define mz(array) memset(array, 0, sizeof(array))
16 #define minf(array) memset(array, 0x3f, sizeof(array))
17 #define REP(i,n) for(int i=0;i<(n);i++)
18 #define FOR(i,x,n) for(int i=(x);i<=(n);i++)
19 #define RD(x) scanf("%d",&x)
20 #define RD2(x,y) scanf("%d%d",&x,&y)
21 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
22 #define WN(x) printf("%d\n",x);
23 #define RE  freopen("D.in","r",stdin)
24 #define WE  freopen("1.out","w",stdout)
25 
26 bool a[1111111];
27 
28 
29 int main() {
30     int x,i,n,j;
31     RD(n);
32     int n2=n/2;
33     for(x=2; x<=n2; x++) {
34         if(a[x]==0) {
35             for(i=x+x; i<=n; i+=x) {
36                 a[i]=1;
37             }
38         }
39     }
40     for(x=4;x<=n2;x++){
41         if(a[x]==1 && a[n-x]==1)break;
42     }
43     printf("%d %d\n",x,n-x);
44     return 0;
45 }

View Code

B - Design Tutorial: Learn from Life

题意：一堆人从1楼坐电梯，电梯一次最多装K人，给出各个人要去的各个楼层，上下电梯时间忽略不计，电梯移动一层需要1秒，求最少需要多少时间才能送完所有人并且电梯回到1楼。

题解：贪心，优先送最高的，空位顺便送低的。

 1 //#pragma comment(linker, "/STACK:102400000,102400000")
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<iostream>
 5 #include<cstring>
 6 #include<algorithm>
 7 #include<cmath>
 8 #include<map>
 9 #include<set>
10 #include<stack>
11 #include<queue>
12 using namespace std;
13 #define ll __int64
14 #define usint unsigned int
15 #define mz(array) memset(array, 0, sizeof(array))
16 #define minf(array) memset(array, 0x3f, sizeof(array))
17 #define REP(i,n) for(int i=0;i<(n);i++)
18 #define FOR(i,x,n) for(int i=(x);i<=(n);i++)
19 #define RD(x) scanf("%d",&x)
20 #define RD2(x,y) scanf("%d%d",&x,&y)
21 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
22 #define WN(x) printf("%d\n",x);
23 #define RE  freopen("D.in","r",stdin)
24 #define WE  freopen("1.out","w",stdout)
25 
26 int a[2222];
27 
28 int main() {
29     int n,k,i,j,x,ans;
30     RD2(n,k);
31     REP(i,n) {
32         RD(x);
33         a[x]++;
34     }
35     x=0;
36     ans=0;
37     for(i=2000;i>=2;i--) {
38         if(a[i]>0) {
39             a[i]-=x;
40             int t=a[i]/k;
41             int tt=t*k;
42             int ttt=tt>=a[i]?t:t+1;
43             ans+=ttt*(i-1)*2;
44             x=ttt*k-a[i];
45             //printf("%d %d %d\n",i,ttt,ans);
46         }
47     }
48     WN(ans);
49     return 0;
50 }

View Code

C - Design Tutorial: Make It Nondeterministic

题意：每个人可以用自己的姓或者名字作为ID，给出各个人的名字，给出一个排名序列，若有一种选择ID的方法可以使排名正确，则输出YES，否则输出NO。

题解：贪心，排名1的人选字典序少的那个，然后都优先选字典序小的，不行了就选大的，再不行就NO。

 1 //#pragma comment(linker, "/STACK:102400000,102400000")
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<iostream>
 5 #include<cstring>
 6 #include<algorithm>
 7 #include<cmath>
 8 #include<map>
 9 #include<set>
10 #include<stack>
11 #include<queue>
12 using namespace std;
13 #define ll __int64
14 #define usint unsigned int
15 #define mz(array) memset(array, 0, sizeof(array))
16 #define minf(array) memset(array, 0x3f, sizeof(array))
17 #define REP(i,n) for(int i=0;i<(n);i++)
18 #define FOR(i,x,n) for(int i=(x);i<=(n);i++)
19 #define RD(x) scanf("%d",&x)
20 #define RD2(x,y) scanf("%d%d",&x,&y)
21 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
22 #define WN(x) printf("%d\n",x);
23 #define RE  freopen("D.in","r",stdin)
24 #define WE  freopen("1.out","w",stdout)
25 
26 char s[111111][2][55];
27 
28 int cmp(char a[55],char b[55]) {
29     int la=strlen(a),lb=strlen(b);
30     int l=min(la,lb);
31     int i;
32     for(i=0; i<l; i++) {
33         if(a[i]>b[i])return 1;
34         if(b[i]>a[i])return -1;
35     }
36     if(la>lb)return 1;
37     if(lb>la)return -1;
38     return 0;
39 }
40 
41 int main() {
42     int n;
43     RD(n);
44     REP(i,n) {
45         scanf(" %s%s",s[i][0],s[i][1]);
46         //printf("%d %s!%s!\n",i,s[i][0],s[i][1]);
47     }
48     char q[55];
49     int x,y=0,z,j;
50     RD(x);
51     y=x-1;
52     if(cmp(s[y][0],s[y][1])==-1) z=0;
53     else z=1;
54     bool flag=1;
55     REP(i,n-1) {
56         RD(x);
57         x--;
58         if(cmp(s[x][0],s[x][1])==-1) j=0;
59         else j=1;
60         if(cmp(s[x][j] , s[y][z])==-1){
61             if(cmp(s[x][j^1], s[y][z])==-1){
62                 flag=0;
63                 break;
64             }else{
65                 y=x;
66                 z=j^1;
67             }
68         }else{
69             y=x;
70             z=j;
71         }
72     }
73     if(flag)puts("YES");
74     else puts("NO");
75 return 0;
76 }

View Code

D - Design Tutorial: Inverse the Problem

题意：给出各个点距离的邻接矩阵，判断这是不是一棵树。

题解：先根据样例特判一些情况，然后最小生成树，我是用可撸斯卡尔，当要加入的边的两个点x和y已经在同一集合，则判它们的距离是否正确。我的方法是找到一个中间点v，使x-v和v-y有路，找到(xv距离) + (vy距离)最小的那个值，当作x到y的距离。这是因为可撸斯卡尔是从小往大加边，肯定有至少一个中间点能使得这个有路。复杂度O(n^3)，略高，姿势不对就要TLE……更好的解法可以看这里：http://www.cnblogs.com/forgot93/p/4000850.html

  1 //#pragma comment(linker, "/STACK:102400000,102400000")
  2 #include<cstdio>
  3 #include<cmath>
  4 #include<iostream>
  5 #include<cstring>
  6 #include<algorithm>
  7 #include<cmath>
  8 #include<map>
  9 #include<set>
 10 #include<stack>
 11 #include<queue>
 12 using namespace std;
 13 #define ll __int64
 14 #define usint unsigned int
 15 #define mz(array) memset(array, 0, sizeof(array))
 16 #define mf1(array) memset(array, -1, sizeof(array))
 17 #define minf(array) memset(array, 0x3f, sizeof(array))
 18 #define REP(i,n) for(int i=0;i<(n);i++)
 19 #define FOR(i,x,n) for(int i=(x);i<=(n);i++)
 20 #define RD(x) scanf("%d",&x)
 21 #define RD2(x,y) scanf("%d%d",&x,&y)
 22 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
 23 #define WN(x) printf("%d\n",x);
 24 #define RE  freopen("D.in","r",stdin)
 25 #define WE  freopen("1.out","w",stdout)
 26 const int inf=2*1e9+7;
 27 const int maxn=2222;
 28 const int maxm=maxn*maxn;
 29 
 30 int a[2222][2222];
 31 int n;
 32 
 33 
 34 int bx[maxm],by[maxm],bz[maxm];
 35 int bn;
 36 inline void add(const int &x,const int &y,const int &z){
 37     bx[bn]=x;
 38     by[bn]=y;
 39     bz[bn]=z;
 40     bn++;
 41 }
 42 
 43 int d[maxm];
 44 bool cmp(const int &x,const int &y){
 45     return bz[x]<bz[y];
 46 }
 47 
 48 int f[maxm];
 49 int gf(const int &x){
 50     if(f[x]==-1)return x;
 51     f[x]=gf(f[x]);
 52     return f[x];
 53 }
 54 
 55 struct Edge{
 56     int next,v,l;
 57 }e[maxm];
 58 int head[maxn],en;
 59 
 60 inline void addedge(const int &x,const int &y,const int &z){
 61     e[en].v=y;
 62     e[en].l=z;
 63     e[en].next=head[x];
 64     head[x]=en;
 65     en++;
 66 }
 67 
 68 int dis[maxn][maxn];
 69 
 70 inline void gank(const int &x,const int &y){
 71     int mi=inf;
 72     for(int i=head[x];i!=-1;i=e[i].next){
 73         if(dis[y][e[i].v]!=-1){
 74             //printf("(%d) %d + %d = %d\n",e[i].v,e[i].l,dis[y][e[i].v],e[i].l + dis[y][e[i].v]);
 75             mi=min(mi , e[i].l + dis[y][e[i].v]);
 76         }
 77     }
 78     dis[y][x]=dis[x][y]=mi;
 79 }
 80 
 81 bool farm(){
 82     int i,j,k,x,y,z;
 83     bn=0;
 84     REP(i,n){
 85         REP(j,n){
 86             if(i!=j){
 87                 if(a[i][j]==0)return 0;
 88             }else{
 89                 if(a[i][j]!=0)return 0;
 90             }
 91         }
 92     }
 93     REP(i,n){
 94         REP(j,i){
 95             if(a[i][j]!=a[j][i])return 0;
 96             add(i,j,a[i][j]);
 97         }
 98     }
 99     REP(i,bn){
100         d[i]=i;
101     }
102     sort(d,d+bn,cmp);
103     mf1(f);
104     //mf1(head);
105     REP(i,n)REP(j,n)dis[i][j]=-1;
106     en=0;
107     mf1(head);
108     REP(i,n)dis[i][i]=0;
109     //en=0;
110     REP(i,bn){
111         int q=d[i];
112         int &x=bx[q],&y=by[q],&z=bz[q];
113         int fx=gf(x),fy=gf(y);
114         if(fx!=fy){
115             addedge(x,y,z);
116             addedge(y,x,z);
117             dis[x][y]=z;
118             dis[y][x]=z;
119             f[fx]=fy;
120         }else{
121             if(dis[x][y]==-1) gank(x,y);
122             //printf("dis[%d][%d]=%d , z=%d\n",x,y,dis[x][y],z);
123             if(dis[x][y]!=z)return 0;
124         }
125     }
126     return 1;
127 }
128 
129 
130 
131 int main() {
132     int i;
133     RD(n);
134     REP(i,n)REP(j,n) RD(a[i][j]);
135     if(farm())puts("YES");
136     else puts("NO");
137     return 0;
138 }

View Code

posted @ 2014-10-03 10:56 带鱼Yuiffy 阅读(251) 评论(0) 编辑收藏举报

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