hdu1029 技巧

/**************************************
*Author :jibancanyang
*Created Time : 二 4/ 5 22:00:39 2016
*题目类型:很好的利用了过半的特性,记录当前最大频率数和他的频率.
***************************************/

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <cmath>#include <cstdlib>#include <ctime>#include <stack>using namespace std;typedef pair<int, int> pii;typedef long long ll;typedef unsigned long long ull;vector<int> vi;#define xx first#define yy second#define sa(n) scanf("%d", &(n))#define rep(i, a, n) for (int i = a; i < n; i++)#define vep(c) for(decltype((c).begin() ) it = (c).begin(); it != (c).end(); it++) const int mod = int(1e9) + 7, INF = 0x3fffffff, maxn = 1e5 + 12;int main(void){#ifdef LOCAL    //freopen("in.txt", "r", stdin);    //freopen("out.txt", "w", stdout);#endif    cin.sync_with_stdio(false);    int n;    while (cin >> n) {    int key = -123433, cnt = 0;        rep(i, 0, n) {            int x;            cin >> x;            if (x != key) {                if (cnt == 0) key = x, cnt++;                else cnt--;            } else cnt++;        }        cout << key << endl;    }    return 0;}Status API Training Shop Blog About