Codeforces Educational Codeforces Round 5 E. Sum of Remainders 数学

E. Sum of Remainders

题目连接：

http://www.codeforces.com/contest/616/problem/E

Description

The only line contains two integers n, m (1 ≤ n, m ≤ 1013) — the parameters of the sum.

input

The only line contains two integers n, m (1 ≤ n, m ≤ 1013) — the parameters of the sum.

Output

Print integer s — the value of the required sum modulo 109 + 7.

Sample Input

3 4

Sample Output

4

Hint

题意

给你n,m，让你输出n%1+n%2+n%3+....+n%m

答案需要mod1e9+7

题解：

n%i = n - i * [n/i]

所以 sigma(n%i) = n * m - sigma(i * n/i)

在一段区间内，n/i都是不会变化的，所以利用这一点，我们就可以暴力sqrt去处理就好了

但是有一些区间我们最后并没有处理到，怎么办？

直接再暴力一发去处理就好了

代码

#include<bits/stdc++.h>
using namespace std;

const int mod = 1e9 + 7;
int main()
{
    long long n,m;
    scanf("%lld%lld",&n,&m);
    long long ans = (n%mod)*(m%mod)%mod;
    long long S = sqrt(n+5);
    long long k = m+1;
    for(int i=1;i<=min(n,S);i++)
    {
        long long r = n / i;
        long long l = n / (i+1)+1;
        r = min(r,m);
        if(l>r)continue;
        k = min(k,l);
        long long s1 = (l+r);
        long long s2 = (r-l+1);
        if(s1%2==0)
            s1/=2LL;
        else
            s2/=2LL;
        long long K = (s1%mod)*(s2%mod)%mod;
        K = (i * K)%mod;
        ans-=K;
        ans%=mod;
        if(ans<0)ans+=mod;
    }
    for(int i=1;i<k;i++)
    {
        ans-=(n-n%i);
        ans%=mod;
        if(ans<mod)ans+=mod;
    }
    cout<<ans%mod<<endl;
}