[CQOI2007] 余数求和

(sum_{i=1}^n;k;mod;i)

Solution

(sum_{i=1}^n;k;mod;i\=sum_{i=1}^n(k-ilfloor{frac{k}{i}} floor)\=k imes n-sum_{i=1}^nilfloor{frac{k}{i}} floor)

至于后面那项，整除分块即可

#include <bits/stdc++.h>
using namespace std;
#define int long long

signed main() {
    int n,k,l=1,r,ans=0;
    cin>>n>>k;
    while(l<=min(n,k)) {
        r=min(n,k/(k/l));
        ans+=(l+r)*(r-l+1)*(k/l)/2;
        l=r+1;
    }
    cout<<k*n-ans;
}