【CQOI2007】【BZOJ1257】余数之和

codes

#include<iostream>
using namespace std;
typedef long long LL;
int main(){
    int n, k;
    cin>>n>>k;
    LL ans = 0;
    if(n > k)ans=(LL)k*(n-k), n=k-1;
    for(int i = 1; i <= n; i++){
        int t = k/i, r = (t?k/t:n);
        if(r > n)r = n;
        ans += (LL)k*(r-i+1)-(LL)t*(i+r)*(r-i+1)/2;
        i = r;
    }
    cout<<ans<<'\n';
    return 0;
}