Codeforces 616E Sum of Remainders 【数学分块】

题目链接：Codeforces 616E Sum of Remainders

E. Sum of Remainders

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

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.

Examples

input

3 4

output

4

input

4 4

output

1

input

1 1

output

0

题意：求解∑mi=1nmodi。

思路：我们化简nmodi=n−n/i∗i。

这样原式 =n∗m−∑mi=1(n/i∗i)

首先m<=n√时，直接暴力，对于剩下的，我们可以分块来搞，总会存在一段连续的值使得n/i为定值，那么我们找到这个区间就可以了。不过貌似在除以2的时候，会出问题，我直接上逆元就过了。

AC代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#define INF 0x7fffffff
using namespace std;
typedef long long LL;
const int MAXN = 1e6 + 10;
const int MOD = 1e9 + 7;
void add(LL &x, LL y) {x += y; x %= MOD;}
LL pow_mod(LL a, LL n) {
LL ans = 1LL;
while(n) {
if(n & 1)
ans = ans * a % MOD;
a = a * a % MOD;
n >>= 1;
}
return ans;
}
int main()
{
LL n, m; scanf("%lld%lld", &n, &m);
//    LL s = 0;
//    for(LL i = 1; i <= m; i++) {
//        add(s, n % i % MOD);
//    }
//    cout << s << endl;
LL ans = n % MOD * (m % MOD) % MOD;
LL nn = sqrt(n);
LL ans1 = 0;
for(LL i = 1; i <= min(m, nn); i++) {
add(ans1, n / i % MOD * i % MOD);
}
//cout << ans1 << endl;
if(m > nn) {
if(nn * nn == n) nn--;
for(LL i = 1; i <= nn; i++) {
LL r = min(m, n / i); LL l = n / (i + 1) + 1;
if(l > r || r <= nn) continue;
//cout << l << ' ' << r << endl;
add(ans1, (l + r) % MOD * ((r - l + 1) % MOD) % MOD * pow_mod(2, MOD-2) % MOD * i % MOD);
}
}
printf("%lld\n", ((ans - ans1) % MOD + MOD) % MOD);
return 0;
}