CodeChef Sereja and GCD
Sereja and GCD |
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, ..., AN, with 1 ≤ Ai ≤ 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。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#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 ; } }
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