Educational Codeforces Round 33 (Rated for Div. 2) E. Counting Arrays

题目链接
题意：给你两个数x,yx,yx,y,让你构造一些长为yyy的数列，让这个数列的累乘为xxx，输出方案数。
思路:考虑对xxx进行质因数分解，设某个质因子PiP_iPi的的幂为kkk,则这个质因子的贡献就相当于把kkk个PiP_iPi放到yyy个盒子中，且盒子可能为空，方案为C(k+y−1,y)C(k+y-1,y)C(k+y−1,y)，然后每个质因子的方案乘在一起即可。最后，因为负号也会出现，但xxx为正，所以就是在yyy个位置上选偶数个位置放负号，方案为2y−12^{y-1}2y−1再乘起来即可。

#include<bits/stdc++.h>#define LL long long
#define fi first
#define se second
#define mp make_pair
#define pb push_backusing namespace std;LL gcd(LL a,LL b){return b?gcd(b,a%b):a;}
LL lcm(LL a,LL b){return a/gcd(a,b)*b;}
LL powmod(LL a,LL b,LL MOD){LL ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
const int N = 2e6 +11;
const LL mod=1e9+7;
LL Fac[N+33],Inv[N+33];
int p[N+33],a[N+33],cnt;
void init(){Fac[0]=1;for(int i=1;i<=N;i++)Fac[i]=(Fac[i-1]*i)%mod;Inv[N]=powmod(Fac[N],mod-2,mod);for(int i=N-1;i>=1;i--)Inv[i]=(Inv[i+1]*(i+1))%mod;Inv[0]=1;
}
void P(){for(int i=2;i<N;i++){if(!p[i])a[++cnt]=i;for(int j=1;j<=cnt&&1ll*a[j]*i<N;j++){p[a[j]*i]=1;if(i%a[j]==0)break;}}
}
LL C(int x,int y){return 1ll*Fac[x]*Inv[y]%mod*Inv[x-y]%mod;
}
int main(){ios::sync_with_stdio(false);init();int t;P();for(cin>>t;t;t--){int x,y;cin>>x>>y;LL ans=1;for(int i=1;i<=cnt&&1ll*a[i]*a[i]<=x;i++){if(x%a[i]==0){int res=0;while(x%a[i]==0)res++,x/=a[i];ans=ans*C(res+y-1,y-1);ans%=mod;}}if(x>1)ans=ans*C(y,y-1)%mod;cout<<ans*powmod(2,y-1,mod)%mod<<endl;}return 0;
}

转载于:https://www.cnblogs.com/pubgoso/p/10759709.html

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