Educational Codeforces Round 5-E. Sum of Remainders

原题链接

E. Sum of Remainders

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard 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

这道题分三步来做:

1.当m > n时　ans += n * (m - n); m = n - 1;

2.当m < n && m > 1e6时,分块取模,比如 m > n / 2那么(n / 2, m]的取模为等差数列求和 = (n % (n/2+1) + n % m) * (m - n + 1)/ 2; m = n / 2;以此类推

3.当 m <= 1e6时for(i = 2; i <= m; i++) ans += n % i;

#include <bits/stdc++.h>
#define maxn 200005
#define MOD 1000000007
using namespace std;
typedef long long ll;

int main(){

// freopen("in.txt", "r", stdin);
ll n, m, ans = 0;

scanf("%I64d%I64d", &n, &m);
if(m > n){
ans += (m - n) % MOD * (n % MOD) % MOD;
}
m = min(n - 1, m);
int cnt = 2;

while(m > 1000000){
ll d = n / cnt + 1;
if(m >= d){
ll p1 = n % d;
ll p2 = n % m;
ans += (p1 + p2) % MOD * ((m - d + 1) % MOD) % MOD * 500000004 % MOD;//2 * 500000004 % MOD == 1ans %= MOD;
if(n % cnt == 0)
m = n / cnt - 1;
else
m = n / cnt;
}
cnt++;
}
for(int i = 2; i <= m; i++){
ans += n % i;
ans %= MOD;
}
cout << ans << endl;

return 0;
}