余数求和

余数求和

 

 

 

 

 题目分析：

\(G\left ( n,m\right ) = \sum_{i=1}^{n} k mod i = \sum_{i=1}^{n}k-\left \lfloor k/i\right \rfloor*i = n*k-\sum_{i=1}^{n}\left \lfloor k/i\right \rfloor*i\)

 AC_Code:

#include <iostream>
#include <cstdio>
#include <algorithm>
typedef long long ll;
using namespace std;

ll n,k;
int main()
{
    scanf("%lld%lld",&n,&k);
    ll ans=n*k;

    for(ll l=1,r;l<=n;l=r+1){
        if( k/l!=0 ) r=min(k/(k/l),n);
        else r=n;
        ans -= (k/l)*(r-l+1)*(l+r)/2;
    }

    printf("%lld\n",ans);
    return 0;
}