Codeforces 616E

616E Sum of Remainders

Calculate the value of the sum: n mod 1 + n mod 2 + n mod 3 + … + n mod m. As the result can be very large, you should print the value modulo 109 + 7 (the remainder when divided by 109 + 7).

The modulo operator a mod b stands for the remainder after dividing a by b. For example 10 mod 3 = 1.
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 test(s)
Input

3 4

Output

4

Input

4 4

Output

1

Input

1 1

Output

0

• 思路：
  1.对mod的理解与转化: n%mod = n - n/mod*mod;
  现在答案就变成了 n*m - Σ(1->m）（n/i*i)，乍一看认为没什么优化，还是要求后面的从1 到m的Σ(n/i*i);这就是优化。现在就需要用到对整除的理解的~~ 对于n/x = n/i;这样的x的区间边界时什么？ 经分析知:[i,n/(n/i)];却这里得到一个很有趣的结论，对于n/(n/(n/i)) = n/i;

  2.然后就是mod运算了，注意在同一个n/i下的区间内数的和(l+r)/2时，为了不爆范围，先用偶数除以二，之后在相乘；
  3.举个例子对下面程序的运行：
  比如n = 15,m = 13;
  则[1,1] ==>是第一个计算的，对于任意的x ,l <= x <= r,n/x = n/i;
  [2,2],[3,3],[4,5],[6,7],[8,13];//区间距离越来越大；这里是优化的对象；
  之后 l = r,再++l;就是对于每一个n/i相等的区间，直接利用等差数列求完，下次求解的时候，l = r+1;即区间不会覆盖；所以对于1e13范围内的数据，还是可以在436ms内求解完毕；如上面最后一次是8~13

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod = 1e9+7;
int add(ll a, ll b) { return (a+b)%mod; }
int sub(ll a, ll b) { return ((a-b)%mod + mod)%mod; }
int mult(ll a, ll b) { return ((a%mod) * (b%mod))%mod; }

int main()
{
    ll i,n,m;
    cin>>n>>m;
    int ans = mult(n,m),sum = 0;
    m = min(n,m);
    for(i = 1;i <= m;i++){// l = i,r = n/(n/i);while:l <= x <= r ;n/x = n/i;
        ll r = n/(n/i);
        r = min(r,m);
        ll sm = i + r,nm = r - i + 1;
        if(sm&1) sm = mult(sm,nm/2);
        else sm = mult(sm/2,nm);
        sum = add(sum,mult(sm,n/i));
        i = r;
    }
    ans = sub(ans,sum);
    cout<<ans;
}