CodeChef Sereja and GCD-优快云博客

Sereja and GCD

Problem code: SEAGCD

All submissions for this problem are available.

Read problems statements in Mandarin Chinese and Russian.

In this problem Sereja is interested in the number of arrays of integers, A₁, A₂, ..., A_N, with 1 ≤ A_i ≤ M, such that the greatest common divisor of all of its elements is equal to a given integer D.

Find the sum of answers to this problem with D = L, D = L+1, ..., D = R, modulo 10⁹+7.

Input

The first line of the input contains an integer T - the number of test cases. T tests follow, each containing a single line with the values of N, M, L, R.

Output

For each test case output the required sum, modulo 10⁹+7.

Constraints

1 ≤ T ≤ 10
1 ≤ L ≤ R ≤ M

Subtasks

Subtask #1: 1 ≤ N, M ≤ 10 (10 points)
Subtask #2: 1 ≤ N, M ≤ 1000 (30 points)
Subtask #3: 1 ≤ N, M ≤ 10⁷ (60 points)

Example

Input:
2
5 5 1 5
5 5 4 5

Output:
3125
2

对于一个 d , 1~m中有 m/d个数是 d的倍数，自然就是有 (m/d)^n种排列方法。

然而，这些排列当中，元素必须包含 d , 只要它减去那些只包含它的倍数的序列即可得出结果。

#include <iostream>
#include <cstdio>
using namespace std ;
typedef long long LL;
const int mod = 1e9+7;
const int N = 10000100;
LL A[N] ;
LL q_pow( LL a , LL b ) {
    LL res = 1 ;
    while( b ) {
        if( b&1 ) res =  res * a  % mod ;
        a = a * a % mod ;
        b >>= 1;
    }
    return res ;
}
int main() {
//    freopen("in.txt","r",stdin);
    int _ ; cin >> _ ;
    while( _-- ) {
        LL n , m , l , r ;
        cin >> n >> m >> l >> r ;
        for( int d = m ; d >= l ; --d ) {
            if( d == m || m/d != m/(d+1) ) A[d] = q_pow( m/d , n );
            else A[d] = A[d+1] ;
        }
        for( int i = m ; i >= l ; --i ) {
            for( int j = i + i ; j <= m ; j += i ) {
                A[i] =( ( A[i] - A[j] ) % mod + mod ) % mod ;
            }
        }
        LL ans = 0 ;
        for( int i = l ; i <= r ; ++i )
            ans = ( ans + A[i] ) % mod ;
        cout << ans << endl ;
    }
}