[CF900D] Unusual Sequences - 组合,容斥,dp
Description
输入x,y,求有多少个数列满足其gcd为x,和为y。
Solution
令 m=y/x,则答案等于 gcd=1,sum=m 的数列个数
设 (f[i]) 表示 sum=i 的数列个数
设 (g[i]) 表示 sum|i 的数列个数
则 (g[i]=sum_{d|i} f[d])
(g[i]) 是容易求出的,根据隔板法,(g[i]=2^{i-1})
求 (f[i]) 有两种方法,一种是手动容斥,一种是莫比乌斯反演
这里我们用手动容斥,枚举 gcd=d,然后删去 f(n/d)
即 (f[i] = g[i] - sum_{d|i, d<i} f[i/d])
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int mod = 1e9 + 7;
namespace math_mod
{
int c__[5005][5005], fac__[3000005];
int qpow(int p, int q)
{
return (q & 1 ? p : 1) * (q ? qpow(p * p % mod, q / 2) : 1) % mod;
}
int inv(int p)
{
return qpow(p, mod - 2);
}
int fac(int p)
{
if (p <= 3000000)
return fac__[p];
if (p == 0)
return 1;
return p * fac(p - 1) % mod;
}
int __fac(int p)
{
return fac(p);
}
int ncr(int n, int r)
{
if (r < 0 || r > n)
return 0;
return fac(n) * inv(fac(r)) % mod * inv(fac(n - r)) % mod;
}
void math_presolve()
{
fac__[0] = 1;
for (int i = 1; i <= 3000000; i++)
{
fac__[i] = fac__[i - 1] * i % mod;
}
for (int i = 0; i <= 5000; i++)
{
c__[i][0] = c__[i][i] = 1;
for (int j = 1; j < i; j++)
c__[i][j] = c__[i - 1][j] + c__[i - 1][j - 1], c__[i][j] %= mod;
}
}
int __c(int n, int r)
{
if (r < 0 || r > n)
return 0;
if (n > 5000)
return ncr(n, r);
return c__[n][r];
}
}
using namespace math_mod;
int m;
map<int, int> f;
int solve(int x)
{
if (f.find(x) != f.end())
return f[x];
int ans = qpow(2, x - 1);
vector<int> fac;
for (int i = 2; i * i <= x; i++)
if (x % i == 0)
{
fac.push_back(i);
if (i * i != x)
fac.push_back(x / i);
}
fac.push_back(1);
for (auto i : fac)
ans = (ans - solve(i) + mod) % mod;
return f[x] = ans;
}
signed main()
{
ios::sync_with_stdio(false);
int x, y;
cin >> x >> y;
m = y / x;
f[1] = 1;
if (y % x)
cout << 0 << endl;
else
cout << solve(m) << endl;
}