题目:
Given a number sequence b 1,b 2…b n.
Please count how many number sequences a 1,a 2,...,a n satisfy the condition that a1*a 2*...*a n=b 1*b 2*…*b n (a i>1).
Input
For each test case, the first line contains an integer n(1<=n<=20). The second line contains n integers which indicate b 1, b 2,...,b n(1<b i<=1000000, b 1*b 2*…*b n<=10 25).
2 3 4
4
For the sample input, P=3*4=12. Here are the number sequences that satisfy the condition: 2 6 3 4 4 3 6 2
思路:把每个质因数出现的次数统计出来,假设质数pi,出现的次数为m次,那么就相当于把m个东西放入n个容器中,但是不允许都为空,如果不管是否为空,那么选择的方法有c(n+m-1,n-1)种,然后算出总的方法数,用容斥定理减去不合法的方法数。
代码:
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<algorithm>
#include<ctime>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<list>
#include<numeric>
using namespace std;
#define LL long long
#define ULL unsigned long long
#define INF 1000000000000
#define mm(a,b) memset(a,b,sizeof(a))
#define PP puts("*********************");
template<class T> T f_abs(T a){ return a > 0 ? a : -a; }
template<class T> T gcd(T a, T b){ return b ? gcd(b, a%b) : a; }
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
// 0x3f3f3f3f3f3f3f3f
// 0x3f3f3f3f
const LL MOD=1e9+7;
const int maxn=1e6+50;
bool isprime[maxn];
int prime[maxn],tol;
void make_prime(int n){
for(int i=0;i<=n;i++)
isprime[i]=true;
isprime[0]=isprime[1]=false;
tol=0;
for(int i=2;i<=n;i++){
if(isprime[i])
prime[tol++]=i;
for(int j=0;j<tol;j++){
if(i*prime[j]<=n)
isprime[i*prime[j]]=false;
else
break;
if(i%prime[j]==0)
break;
}
}
}
vector<int> fac;
int num[405],b[25];
LL c[1005][1005];
LL C(int n,int m){
return c[n+m-1][n-1];
}
int main(){
int n;
make_prime(maxn-50);
for(int i=0;i<=1000;i++)
c[i][0]=c[i][i]=1;
for(int i=1;i<=1000;i++)
for(int j=1;j<i;j++)
c[i][j]=(c[i-1][j-1]+c[i-1][j])%MOD;
while(~scanf("%d",&n)){
fac.clear();
for(int i=0;i<n;i++)
scanf("%d",&b[i]);
for(int i=0;i<n;i++){
int x=b[i];
for(int j=0;j<tol&&prime[j]*prime[j]<=b[i];j++){
while(x%prime[j]==0){
fac.push_back(prime[j]);
x/=prime[j];
}
}
if(x>1) fac.push_back(x);
}
sort(fac.begin(),fac.end());
mm(num,0);
int cnt=0;
for(int i=0;i<fac.size();i++){
if(i==0)
num[++cnt]=1;
else{
if(fac[i]==fac[i-1]) num[cnt]++;
else num[++cnt]=1;
}
}
LL ans=1;
for(int i=1;i<=cnt;i++)
ans=(ans*C(n,num[i]))%MOD;
for(int i=1;i<=n-1;i++){//枚举空容器的个数
LL tmp=c[n][i];
for(int j=1;j<=cnt;j++)
tmp=(tmp*C(n-i,num[j]))%MOD;
// printf("%d %lld\n",i,tmp);
if(i%2==1){
ans=(ans-tmp)%MOD;
ans=(ans+MOD)%MOD;
}
else
ans=(ans+tmp)%MOD;
}
printf("%lld\n",ans);
}
return 0;
}