CodeChef November Lunchtime 2013 Lucy and the Number Game(简单题)

n个人报数找到只有一个报过且最小的数。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#define INF 0x7fffffff
#define LEN 1000100
using namespace std;

struct P{
    char name[50];
    int num;
}man[LEN];

bool cmp(struct P a, struct P b){
    return a.num<b.num;
}

int main()
{
//    freopen("in.txt", "r", stdin);

    int T, n;
    scanf("%d", &T);
    while(T--){
        scanf("%d", &n);
        for(int i=1; i<=n; i++){
            scanf("%s%d", man[i].name, &man[i].num);
        }
        sort(man+1, man+n+1, cmp);
        int ans = -1, loc = man[1].num, cnt=1;
        for(int i=2; i<=n; i++){
            if(man[i].num == loc){
                cnt++;
            }
            else {
                if(cnt == 1){
                    ans = i-1;
                    break;
                }else{
                    loc = man[i].num;
                    cnt = 1;
                }
            }
        }
        if(cnt == 1 && loc == man[n].num) ans = n;
        if(ans!=-1)printf("%s
", man[ans].name);
        else printf("Nobody wins.
");
    }
    return 0;
}