HDU 1029 Ignatius and the Princess IV （思路）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1029大意：从一个N元素的序列（N为奇数）中找出出现次数大于（N+1）/ 2 的元素。思路：设答案为result。按照序列依次扫描，先把数组中第一个元素赋值给result，增加个计数器，cnt = 1；然后向右扫描，如果跟result相同，则cnt++，不同，那么cnt --。当cnt为0时，result换为当前的数。最后的result即为答案。#include<cstdio>#include<cstring>#include<string>#include<cctype>#include<iostream>#include<set>#include<map>#include<cmath>#include<sstream>#include<vector>#include<stack>#include<queue>#include<algorithm>#define fin(a) freopen("a.txt","r",stdin)#define fout(a) freopen("a.txt","w",stdout)#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1typedef long long LL;using namespace std;const int maxn = 1e6 + 10;int result, cnt;int main() {int n, x;while(scanf("%d", &n) == 1) {cnt = 0;while(n--) {scanf("%d", &x);if(cnt == 0) {result = x;cnt = 1;}else if(result == x) cnt++;else cnt--;}printf("%d\n", result);}return 0;}