Luo Gu P2261 [CQOI2007] remainder sum

topic

Number theory, considering the original title to the formula derived \ (i \% j = i- (i / j) * j) \)

原题求\(\sum_{i=1}^{i=n}(k\%i)=\sum_{i=1}^{i=n}(k-i*(k/i))=k*n-\sum_{i=1}^{i=n}(i*(k/i))\)

Thus the original title translates into fast seek \ (\ sum_. 1 = {I}} ^ {n-I = (I * (K / I)) \) , \ (K / I \) is constant over time therefore considered number theory block.

#include <bits/stdc++.h>
#define int long long
using namespace std;
int n, k, ans;
signed main()
{
    scanf("%lld%lld", &n, &k);
    ans = n * k;
    int l, r;
    for (l = 1, r; l <= n; l = r + 1)//l即i，r是满足k/l与k/r相等的数的最右边一位，易得k/(k/l)和n的最小值即为r，然后再用l，r的区间长度乘公差为1的等差数列的和，这样的时间复杂度是O(sqrt(n))的
    {
        if (k / l == 0) r = n;
        else r = min(n, k / (k / l));
        ans -= (k / l) * (r - l + 1) * (l + r) / 2 ;
    }
    printf("%lld", ans);
    return 0;
}