当水军刚好是一半时的问题

见转载帖子的回复部分：

http://blog.csdn.net/v_july_v/article/details/6890054

#include<iostream>#include <stdio.h>using namespace std;   int find( int *a, int n ){intcandidate;intntimes, i;for( i = ntimes = 0; i < n; i++ ){if( ntimes == 0 ){candidate = a[i], ntimes = 1;       } else{if( a[i] == candidate )ntimes++;elsentimes--;}intctimes;intcandidate2 = a[n - 1];for( i = ctimes = 0; i < n; i++ ){if( a[i] == candidate )ctimes++;}return ctimes == n /2 ? candidate : candidate2;}}        int main()  {  int a[]={ 1,1,3,4,1,1,2,2 };          cout<<find(a,sizeof(a)/sizeof(int))<<endl;     //  int a[6]={1,0,2,1,2,1};    //  cout<<Find(a,6)<<endl;     }