Codeforces 900D-Unusual Sequences

Unusual Sequences
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Count the number of distinct sequences a1, a2, ..., an (1 ≤ ai) consisting of positive integers such that gcd(a1, a2, ..., an) = x and . As this number could be large, print the answer modulo 109 + 7.

gcd here means the greatest common divisor.

Input

The only line contains two positive integers x and y (1 ≤ x, y ≤ 109).

Output

Print the number of such sequences modulo 109 + 7.

Examples
input
3 9
output
3
input
5 8
output
0
Note

There are three suitable sequences in the first test: (3, 3, 3)(3, 6)(6, 3).

There are no suitable sequences in the second test.


题意:问能构造出多少个序列满足,序列中所有元素的gcd为x,序列中所有元素的和为y

解题思路:容斥,分别求出gcd为x,2*x……的数量,然后减去多余即可


#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;
const LL mod = 1e9 + 7;
mapmp;

LL qpow(int k)
{
	LL ans = 1, x = 2;
	while (k)
	{
		if (k & 1) (ans *= x) %= mod;
		(x *= x) %= mod;
		k >>= 1;
	}
	return ans;
}

LL solve(int x)
{
	if (x == 1) return 1;
	if (mp[x]) return mp[x];
	mp[x] = qpow(x - 1);//隔板法可以计算出把一个数x分解成可以由几个数组成的序列有2^(x−1)个
	for (int i = 2; i*i <= x; i++)
	{
		if (x%i == 0)
		{
			mp[x] = (mp[x] - solve(x / i) + mod) % mod;
			if (i != x / i) mp[x] = (mp[x] - solve(i) + mod) % mod;
		}
	}
	mp[x] = (mp[x] - solve(1) + mod) % mod;
	return mp[x];
}

int main()
{
	int x, y;
	while (~scanf("%d%d", &x, &y))
	{
		if (y%x != 0) { printf("0\n"); continue; }
		mp.clear();
		printf("%lld\n", solve(y / x));
	}
	return 0;
}

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