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  • MemSQL Start[c]UP 2.0

    A. Eevee http://codeforces.com/contest/452/problem/A

    字符串水题

     1 #include<cstdio>
     2 #include<cstring>
     3 using namespace std;
     4 const int M=16;
     5 char a[16][M]={"vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"};
     6 char b[M];
     7 int main(){
     8     int n;
     9     while(~scanf("%d",&n)){
    10         scanf("%s",b);
    11         for(int i=0;i<8;i++){
    12             int la=strlen(a[i]);
    13             if(la==n){
    14                 bool flag=true;
    15                 for(int j=0;j<n;j++){
    16                     if(b[j]!='.'&&b[j]!=a[i][j]){
    17                         flag=false;
    18                         break;
    19                     }
    20                 }
    21                 if(flag){
    22                     puts(a[i]);
    23                     break;
    24                 }
    25             }
    26         }
    27     }
    28     return 0;
    29 }
    View Code

     B. 4-point polyline http://codeforces.com/contest/452/problem/B

    暴力几种画法

     1 #include<cstdio>
     2 #include<cstdlib>
     3 #include<cstring>
     4 #include<cmath>
     5 #include<cctype>
     6 #include<iostream>
     7 #include<algorithm>
     8 #include<queue>
     9 #include<stack>
    10 #include<vector>
    11 #include<map>
    12 #include<set>
    13 #define mt(a,b) memset(a,b,sizeof(a))
    14 using namespace std;
    15 typedef __int64 LL;
    16 const int inf=0x7fffffff;
    17 const int M=100;
    18 struct point{
    19     int x,y;
    20 };
    21 struct G{
    22     point p[4];
    23     int val;
    24     friend bool operator <(G a,G b){
    25         return a.val>b.val;
    26     }
    27 }g[M];
    28 int dist(point a,point b){
    29     return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
    30 }
    31 int lg;
    32 void add(int id,int x,int y){
    33     g[lg].p[id].x=x;
    34     g[lg].p[id].y=y;
    35 }
    36 int main(){
    37     int n,m;
    38     while(~scanf("%d%d",&n,&m)){
    39         if(n==0){
    40             printf("0 1
    ");
    41             printf("0 %d
    ",m);
    42             printf("0 0
    ");
    43             printf("0 %d
    ",m-1);
    44             continue;
    45         }
    46         if(m==0){
    47             printf("1 0
    ");
    48             printf("%d 0
    ",n);
    49             printf("0 0
    ");
    50             printf("%d 0
    ",n-1);
    51             continue;
    52         }
    53         lg=0;
    54         add(0,0,0);
    55         add(1,n,m);
    56         add(2,0,m);
    57         add(3,n,0);
    58         lg++;
    59         add(0,0,0);
    60         add(1,n,m);
    61         add(2,n,0);
    62         add(3,0,m);
    63         lg++;
    64         add(0,1,0);
    65         add(1,n,m);
    66         add(2,0,0);
    67         add(3,n-1,m);
    68         lg++;
    69         add(0,0,1);
    70         add(1,n,m);
    71         add(2,0,0);
    72         add(3,n,m-1);
    73         lg++;
    74         add(0,1,1);
    75         add(1,n,m);
    76         add(2,0,0);
    77         add(3,n-1,m-1);
    78         lg++;
    79         for(int i=0;i<lg;i++){
    80             g[i].val=0;
    81             for(int j=0;j<3;j++){
    82                 g[i].val+=dist(g[i].p[j],g[i].p[j+1]);
    83             }
    84         }
    85         sort(g,g+lg);
    86         for(int i=0;i<4;i++){
    87             printf("%d %d
    ",g[0].p[i].x,g[0].p[i].y);
    88         }
    89     }
    90     return 0;
    91 }
    View Code
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  • 原文地址:https://www.cnblogs.com/gaolzzxin/p/3872544.html
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