The contest is still in progress this moment. How excited it is! You, smart programmer, must have AC some problems today. "Can I get copper medal, silver medal, or even golden medal?" Oh, ha-ha! You must be considering this question. And now, the last problem of this contest comes.
Give you all submitting data in the contest, and tell you the number of golden medals, silver medals and copper medals; your task is to output someone's contest result.
Easy? Of course! I t is the reason that I designed the problem.
When you have completed this contest, please remember that sentence〃 Accepted today?〃兒
A test case starting with 0 0 0 0 0 terminates input and this test case should not to be processed.
Accepted today? I've got a golden medal :)
Accepted today? I've got a silver medal :)
Accepted today? I've got a copper medal :)
Accepted today? I've got an honor mentioned :)
Note:
You will get an honor mentioned if you can't get copper medal, silver medal or golden medal.
简单排序
#include <stdio.h> #include <algorithm> using namespace std; struct AC { int num,h,m,s; int no; int k;//k=1金,2银,3铜,4,安慰奖 } a[200]; int cmp(AC x,AC y) { if(x.num!=y.num) return x.num>y.num; if(x.h!=y.h) return x.h<y.h; if(x.m!=y.m) return x.m<y.m; return x.s<y.s; } int main() { int n,g,s,c,m,i; while(~scanf("%d%d%d%d%d",&n,&g,&s,&c,&m)) { if(n+g+s+m+c == 0) break; for(i = 0; i<n; i++) { a[i].no = i+1; scanf("%d %d:%d:%d",&a[i].num,&a[i].h,&a[i].m,&a[i].s); } sort(a,a+n,cmp); //printf("====\n"); for(i = 0; i<n; i++) { if(g) { a[i].k = 1; g--; } else if(s) { a[i].k = 2; s--; } else if(c) { a[i].k = 3; c--; } else { a[i].k = 4; } } for(i = 0; i<n; i++) { if(a[i].no == m) { if(a[i].k == 1) printf("Accepted today? I've got a golden medal :)\n"); else if(a[i].k == 2) printf("Accepted today? I've got a silver medal :)\n"); else if(a[i].k == 3) printf("Accepted today? I've got a copper medal :)\n"); else if(a[i].k == 4) printf("Accepted today? I've got an honor mentioned :)\n"); break; } } } return 0; }