正题

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题目大意

求${\sum }_{i=1}^{n}k\%i$。

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解题思路

将$k\%i$展开一下，$k-i\ast \lfloor k/i\rfloor$，然后答案就是
${\sum }_{i=1}^{n}k-i\ast \lfloor k/i\rfloor$
将k取出来
$nk-{\sum }_{i=1}^{n}i\ast \lfloor k/i\rfloor$
然后用等差公式分多块计算${\sum }_{i=1}^{n}i\ast \lfloor k/i\rfloor$

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code

#include<cstdio>
#include<algorithm>
using namespace std;
long long n,ans,k;
int main()
{scanf("%lld%lld",&n,&k);ans=n*k;for(int x=1,gx;x<=n;x=gx+1){gx=k/x?min(k/(k/x),n):n;ans-=(k/x)*(x+gx)*(gx-x+1)/2;}printf("%lld",ans);
}