8
7
2015
3

BZOJ 2456: mode

Description

给你一个n个数的数列,其中某个数出现了超过n div 2次即众数,请你找出那个数。

数据范围:  100%的数据,n<=500000,数列中每个数<=maxlongint。

数据范围太大,于是在网上找的了一个比较神奇的找众数的方法:

因为题中规定了众数一定多于二分之一,所以可以每次搜到不同的数都抵消,最后剩下的一定是最多的那个;

代码:


/**************************************************************
    Problem: 2456
    User: 1115825064
    Language: C++
    Result: Accepted
    Time:372 ms
    Memory:804 kb
****************************************************************/
 
 
#include<cstdio>
 
using namespace std;
 
int n,p;
int x,a;
int main()
{
    scanf("%d",&n);
    while(n--)
    {
     
        scanf("%d",&a);
        if(!p)
        {
            p=1;
            x=a;
        }
        else
        {
            if(x==a) p++;
            else p--;
        }
    }
    printf("%d",x);
    return 0;
}
 
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2019年11月03日 09:52

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