[CQOI2007]余数求和

$$k\ mod \ i=k-\left \lfloor \frac{k}{i} \right \rfloor*i$$ $$\sum_{i=1}^{n}k \ mod \ i=\sum_{i=1}^{n}\left ( k-\left \lfloor \frac{k}{i} \right \rfloor*i \right )=nk-\sum_{i=1}^{n}\left ( \left \lfloor \frac{k}{i} \right \rfloor*i \right )$$

分块求解即可

#include<cstdio>
#include<cmath>
#include<algorithm>
#define LL long long
using namespace std;
LL n,k,ans=0;
int main()
{
    scanf("%lld%lld",&n,&k);
    LL sz=sqrt(k);
    for(LL i=1;k/i>sz;i++) ans-=i*(k/i);
    for(LL i=sz;i>=1;i--)
    {
        LL L=k/(i+1)+1,R=min(n,k/i);
        ans-=(L+R)*(R-L+1)/2*i;
    }
    printf("%lld",ans+n*k);
    return 0;
}