2020牛客暑期多校训练营（第四场）B.Basic Gcd Problem

2020牛客暑期多校训练营（第四场）B.Basic Gcd Problem

题目链接

题目描述

As a great ACMer, ZYB is also good at math and number theory.

ZYB constructs a function $f_c(x)$ such that:

Give some positive integer pairs $(n_i,c_i)$, ZYB wants to know $f_{c_i}(n_i)\mathrm{m}\mathrm{o}\mathrm{d}(10^9+7)$.

输入描述:

The input contains multiple test cases. The first line of input contains one integer T $(1\le T\le 10^6)$.
In the following $T$ lines, each line contains two integers $n_i,c_i(1\le n_i,c_i\le 10^6)$ describing one question.

输出描述:

For each test case, output one integer indicating the answer.

示例1

输入

2
3 3
10 5

输出

3
25

数论+找规律，打表就很容易发现：
答案就是 $c^{f(n)}$，$f(n)$ 为 $n$ 的质因子个数(可重复)，然后套一个快速幂就过了~
下面是打表代码:

ll dfs(ll n,ll c){
    if(n==1) return 1;
    ll ans=0;
    for(ll i=1;i<=n-1;i++){
        ans=max(ans,c*dfs(__gcd(i,n),c));
    }
    return ans;
}

求质因子个数我个人认为应该不能暴力,要素筛加质因子分解,这样求是最快的,AC代码如下：

#include
using namespace std;
typedef long long ll;
const ll mod=1e9+7;
const ll N=1e6+10;
ll power(ll a,ll b){return b?power(a*a%mod,b/2)*(b%2?a:1)%mod:1;}
ll prime[N],k,t,n,c;
bool isprime[N];
void Prime(){
    fill(isprime,isprime+N,1);
    k=0;
    prime[1]=0;
    for(ll i=2;i<N;i++){
        if(isprime[i]){
            prime[k++]=i;
            for(ll j=2;i*j<N;j++)
                isprime[i*j]=0;
        }
    }
}

ll solve(ll n){
    ll cnt=0,sum=0;
    for(ll i=0;i<k&&prime[i]*prime[i]<=n;i++){
        if(n%prime[i]==0){
            while(n%prime[i]==0){
                sum++;
                n/=prime[i];
            }
        }
    }
    if(n>1) sum++;
    return sum;
}

int main(){
    Prime();
    scanf("%lld",&t);
    while(t--){
        scanf("%lld%lld",&n,&c);
        printf("%lld\n",power(c,solve(n)));
    }
}