[CQOI2007] 余数求和 - 整除分块

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

Solution

\(\sum_{i=1}^n\;k\;mod\;i\\=\sum_{i=1}^n(k-i\lfloor{\frac{k}{i}}\rfloor)\\=k\times n-\sum_{i=1}^ni\lfloor{\frac{k}{i}}\rfloor\)

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;
}

[CQOI2007] 余数求和 - 整除分块

原文：https://www.cnblogs.com/mollnn/p/12298974.html