CodeForces 689C Mike and Chocolate Thieves （二分最大化最小值）

#include<stdio.h>
#include<iostream>
using namespace std;
int main()
{
    long long  n;
    scanf("%I64d",&n);
    long long left=1,right=1e18,mid,num,m,s;
    long long ans=-1;
    while(left<=right)
    {
        mid=(left+right)>>1;
        num = 0;
        for(long long i=2;i <= 1000000; i++)
        {
            m = i*i*i;
            s=mid/m;
            num+=s;
            if(s==0)break;
        }
        if(num==n)
            ans=mid;
        if(num>=n)
        {
            right=mid-1;
        }
        else
        {
            left=mid+1;
        }
    }
    printf("%I64d\n",ans);
}