洛谷P2397 yyy loves Maths VI (mode) 摩爾投票

求衆數的一種方法

不需要內存，排序

每次記錄上次加的值以及個數，相等個數+1，不相等就-1

因爲衆數大於一半 最壞的情況也就是拿一個衆數取走一個非衆數

最後的結果肯定就是衆數

//#pragma comment (linker, "/STACK:102400000,102400000")
#include<bits/stdc++.h>
#include<stdio.h>
#include<string.h>
#include<string>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<set>
#include<stack>
#include<vector>
#include<map>
#include<queue>
#include<list>
#include<time.h>
#include<bitset>

#define myself i,l,r
#define lson i<<1
#define rson i<<1|1
#define Lson i<<1,l,mid
#define Rson i<<1|1,mid+1,r
#define half (l+r)/2
#define lowbit(x) x&(-x)
#define min4(a, b, c, d) min(min(a,b),min(c,d))
#define min3(x, y, z) min(min(x,y),z)
#define max3(x, y, z) max(max(x,y),z)
#define max4(a, b, c, d) max(max(a,b),max(c,d))
#define pii make_pair
#define pr pair<int,int>
typedef unsigned long long ull;
typedef long long ll;
const int inff = 0x3f3f3f3f;
const long long inFF = 9223372036854775807;
const int dir[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};
const int mdir[8][2] = {0, 1, 0, -1, 1, 0, -1, 0, 1, 1, -1, 1, 1, -1, -1, -1};
const double eps = 1e-10;
const double PI = acos(-1.0);
const double E = 2.718281828459;
using namespace std;
const int mod=1e9+7;
int num,cnt;
int main()
{
    int n,x;
    cnt=0;
    cin>>n;
    while(n--)
    {
        cin>>x;
        if(cnt==0) cnt++,num=x;
        else
        {
            if(num!=x) cnt--;
            else cnt++;
        }
    }
    cout<<num<<endl;
}