[日常摸鱼]bzoj1257余数之和

题意：输入$k,n$，求$sum_{i=1}^n k mod i$

$k mod i=k-i*lfloor frac{k}{i} floor $，$n$个$k$直接求和，后面那个东西像比较套路的分段求和

算k/(k/i)这种东西的时候还要注意判一下分母为0什么的…

#include<cstdio>
typedef long long lint;
lint n,k,ans;
inline lint min(lint a,lint b){return a<b?a:b;}
inline lint getsum(lint i)
{
    return i*(i+1)/2;
}
int main()
{
    scanf("%lld%lld",&n,&k);
    ans=n*k;
    for(register lint i=1,pos;i<=n;i=pos+1)
    {
        pos=(k/i?min(n,k/(k/i)):n);
        ans-=(getsum(pos)-getsum(i-1))*(k/i);
    }
    printf("%lld",ans);
    return 0;
}