A-ace of Aces--zoj

Source: Internet
Author: User

A-Ace of AcesTime limit:$ MS Memory Limit:65536KB 64bit IO Format:%lld & Amp %llu SubmitStatus


There is a mysterious organization called Time-space administrative Bureau (TSAB) in the deep universe that we Hu Mans has not discovered yet. This year, the TSAB decided to elect a outstanding member from its elite troops. The elected guy would 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 the most tickets nominated would be elected as the "Ace of Aces". If there is, or more candidates has the same number of nominations, no one would win.

Tsab determine who'll be the "Ace of Aces".


There is multiple test cases. The first line of input contains an integer indicating the number of the T 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).


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

Sample Input

352 2 2 1 151 1 2 2 31998

Sample Output


Using array storage, find the first second-largest number and compare it!

#include <stdio.h> #include <string.h> #define MAX 2000typedef struct node{int v;int c;} Node;int Main () {int n,m,v[max],t,i,j,max,maxx,k,flag; NODE que[1005];scanf ("%d", &n), for (k=1;k<=n;k++) {flag=0;memset (v,0,sizeof (v)); scanf ("%d", &m); for (I=1; i<=m;i++) {scanf ("%d", &t); v[t]++;} int Len=0;for (i=1;i<max;i++) {if (V[i]) {que[len].c=i;que[len].v=v[i];len++;}} Max=0;for (i=0;i<len;i++) {if (MAX<QUE[I].V) {max=que[i].v;maxx=que[i].c;t=i;}} int Temp=max;int tempx=maxx;que[t].v=-1;max=0;for (i=0;i<len;i++) {if (MAX<QUE[I].V) {Max=que[i].v;maxx=que[i] . C;}} if (max==temp) printf ("nobody\n"); elseprintf ("%d\n", tempx);} return 0;}

A-ace of Aces--zoj

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