ACM刷题之ZOJ————Ace of Aces

Ace of AcesTime Limit: 2 Seconds      Memory Limit: 65536 KBThere is a mysterious organization called Time-Space Administrative Bureau (TSAB) in the deep universe that we humans have not discovered yet. This year, the TSAB decided toelect an outstanding member from its elite troops. The elected guy will be honored with the title of "Ace of Aces".After voting, the TSAB received N valid tickets. On each ticket, there is a number Ai denoting the ID of a candidate. The candidate with themost tickets nominated will be elected as the "Ace of Aces". If there are two or more candidates have the same number of nominations, no one will win.Please write program to help TSAB determine who will be the "Ace of Aces".

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:The first line contains an integer N (1 <= N <= 1000). The next line contains N integers Ai (1 <= Ai <=1000).

Output

For each test case, output the ID of the candidate who will be honored with "Ace of Aces". If no one win the election, output "Nobody" (without quotes) instead.

Sample Input

3
5
2 2 2 1 1
5
1 1 2 2 3
1
998

Sample Output

2
Nobody
998

2015浙江省赛水题

下面是ac代码

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<math.h>#include<algorithm>#include<map>#include<set>#include<queue>#include<string>#include<iostream>using namespace std;#define MID(x,y) ((x+y)>>1)#define CLR(arr,val) memset(arr,val,sizeof(arr))#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);const double PI = acos(-1.0);const int INF = 0x3f3f3f3f;const int N=2e5+7;struct dd{int val,cnt;};dd a[10000];int main(){//freopen("f:/input.txt", "r", stdin);int zu,i,j,k,x,cc,f,maxn,v,q;scanf("%d",&zu);while(zu--){CLR(a,0);scanf("%d",&k);cc=0;for(i=0;i<k;i++){scanf("%d",&x);f=0;for(j=0;j<cc;j++){if(a[j].val==x){f=1;++a[j].cnt;break;}}if(f==0){a[cc].val=x;++a[cc].cnt;++cc;}}maxn=-1;v=-1;for(i=0;i<cc;i++){if(maxn<a[i].cnt){maxn=a[i].cnt;v=a[i].val;}}q=0;for(i=0;i<cc;i++){if(maxn==a[i].cnt)++q;}if(q==1){printf("%d\n",v);}else{printf("Nobody\n");}}}