对于每组数据,给出x和y,求一个长度为y的序列,其乘积为x,允许有负数,求这种序列的个数,对1e9+7取模
首先不考虑正负号,那么我们可以把原题看作把(x)的质因子分配到序列的若干个位置中,那么设(x=prod_ip_i^{k_i}),则对于每个(p_i),都有把(k_i)个小球放到(y)个有标号盒子中,并且盒子可以为空,相当于有(k_i+y)个小球放到(y)个有标号盒子中,每个盒子不能为空,于是用插板法解决,方案数为(egin{pmatrix}k_i+y-1\y-1end{pmatrix})。
然后考虑正负号的问题,我们肯定是每次选择序列中偶数个位置,把他们变为负数,所以有如下式子:
[sum_{i=0}^{lfloorfrac{y}{2}
floor}egin{pmatrix}y\2iend{pmatrix}
]
考虑化简这个式子,那么先考虑(y)是奇数,那么有:
[egin{aligned}sum_{i=0}^{frac{y-1}{2}}egin{pmatrix}y\2iend{pmatrix}\
=sum_{i=0}^frac{y-1}{2}egin{pmatrix}y\iend{pmatrix}end{aligned}]
而我们发现(2^y=sum_{i=0}^yegin{pmatrix}y\iend{pmatrix},egin{pmatrix}y\iend{pmatrix}=egin{pmatrix}y\y-iend{pmatrix}),那么原式就(=2^{y-1})。
然后考虑(y)是偶数,那么有:
[sum_{i=0}^{frac{y}{2}}egin{pmatrix}y\2iend{pmatrix}
]
然后如果(frac{y}{2})是奇数,那么有:
[egin{aligned}2sum_{i=0}^{frac{frac{y}{2}-1}{2}}egin{pmatrix}y\2iend{pmatrix}\=2sum_{i=0}^{frac{y}{2}-1}egin{pmatrix}y-1\iend{pmatrix}end{aligned}\=2(2^{y-2})=2^{y-2}
]
而如果(frac{y}{2})是偶数,那么有:
[egin{aligned}2sum_{i=0}^{frac{frac{y}{2}-2}{2}}egin{pmatrix}y\2iend{pmatrix}end{aligned}+egin{pmatrix}y\frac{y}{2}end{pmatrix}\=2sum_{i=0}^{frac{y}{2}-2}egin{pmatrix}y-1\iend{pmatrix}+egin{pmatrix}y\frac{y}{2}end{pmatrix}\=2(2^{y-2}-egin{pmatrix}y-1\frac{y}{2}-1end{pmatrix})+egin{pmatrix}y-1\frac{y}{2}end{pmatrix}+egin{pmatrix}y-1\frac{y}{2}-1end{pmatrix}\=2^{y-1}+egin{pmatrix}y-1\frac{y}{2}end{pmatrix}-egin{pmatrix}y-1\frac{y}{2}-1end{pmatrix}\=2^{y-1}
]
综上所述可以得到(sum_{i=0}^{lfloorfrac{y}{2} floor}egin{pmatrix}y\2iend{pmatrix}=2^{y-1})。
Code
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
const int N = 2e6;
const int p = 1e9 + 7;
using namespace std;
int T,x,y,fac[N + 5],inv[N + 5],ans;
int mypow(int a,int x){int s = 1;for (;x;x & 1 ? s = 1ll * s * a % p : 0,a = 1ll * a * a % p,x >>= 1);return s;}
int C(int n,int m)
{
return 1ll * fac[n] * inv[n - m] % p * inv[m] % p;
}
int main()
{
scanf("%d",&T);
fac[0] = 1;
for (int i = 1;i <= N;i++)
fac[i] = 1ll * fac[i - 1] * i % p;
inv[1] = 1;
for (int i = 2;i <= N;i++)
inv[i] = 1ll * (p - p / i) * inv[p % i] % p;
inv[0] = 1;
for (int i = 1;i <= N;i++)
inv[i] = 1ll * inv[i - 1] * inv[i] % p;
while (T--)
{
scanf("%d%d",&x,&y);
ans = mypow(2,y - 1);
int xx = x;
for (int i = 2;i * i <= xx;i++)
if (xx % i == 0)
{
int cnt = 0;
while (xx % i == 0)
{
xx /= i;
cnt++;
}
ans = 1ll * ans * C(cnt + y - 1,y - 1) % p;
}
if (xx != 1)
ans = 1ll * ans * y % p;
printf("%d
",ans);
}
return 0;
}