题目链接

题面：

题解：
我们可以知道 f c i ( n i ) = c i k f_{c_i}(n_i)=c_i^k fci​​(ni​)=cik​ 其中 k 是 n 的质因子个数。

一、直接暴力求也没T。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<string>
#include<queue>
#include<bitset>
#include<map>
#include<unordered_map>
#include<set>
#define ui unsigned int
#define ll long long
#define llu unsigned ll
#define ld long double
#define pr make_pair
#define pb push_back
#define lc (cnt<<1)
#define rc (cnt<<1|1)
#define len(x) (t[(x)].r-t[(x)].l+1)
#define tmid ((l+r)>>1)
using namespace std;

const int inf=0x3f3f3f3f;
const ll lnf=0x3f3f3f3f3f3f3f3f;
const double dnf=1e18;
const int mod=1e9+7;
const double eps=1e-8;
const double pi=acos(-1.0);
const int hp=13331;
const int maxn=1000100;
const int maxm=100100;
const int up=100100;


int prime[maxn],cnt;
int ha[maxn];

void Prime(void)
{
    ha[1]=1;
    for(int i=2;i<maxn;i++)
    {
        if(!ha[i])
        {
            ha[i]=i;
            prime[++cnt]=i;
        }
        for(int j=1;j<=cnt&&prime[j]*i<maxn;j++)
        {
            ha[prime[j]*i]=prime[j];
            if(i%prime[j]==0) break;
        }
    }
}

ll only(int x,ll c)
{
    ll ans=1;
    while(x!=1)
        ans=ans*c%mod,x/=ha[x];
    return ans;
}

int main(void)
{
    int tt;
    scanf("%d",&tt);
    Prime();
    while(tt--)
    {
        int n,c;
        scanf("%d%d",&n,&c);
        printf("%lld\n",only(n,c));
    }
    return 0;
}

二、可以O（n）打个表。
也快不了多少嘛，100ms

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<string>
#include<queue>
#include<bitset>
#include<map>
#include<unordered_map>
#include<set>
#define ui unsigned int
#define ll long long
#define llu unsigned ll
#define ld long double
#define pr make_pair
#define pb push_back
#define lc (cnt<<1)
#define rc (cnt<<1|1)
#define len(x) (t[(x)].r-t[(x)].l+1)
#define tmid ((l+r)>>1)
using namespace std;

const int inf=0x3f3f3f3f;
const ll lnf=0x3f3f3f3f3f3f3f3f;
const double dnf=1e18;
const int mod=1e9+7;
const double eps=1e-8;
const double pi=acos(-1.0);
const int hp=13331;
const int maxn=1000100;
const int maxm=100100;
const int up=100100;


int prime[maxn],sum[maxn],cnt;
int ha[maxn];

void Prime(void)
{
    ha[1]=1;
    sum[1]=0;
    for(int i=2;i<maxn;i++)
    {
        if(!ha[i])
        {
            ha[i]=i;
            prime[++cnt]=i;
            sum[i]=1;
        }
        for(int j=1;j<=cnt&&prime[j]*i<maxn;j++)
        {
            ha[prime[j]*i]=prime[j];
            sum[prime[j]*i]=sum[i]+1;
            if(i%prime[j]==0) break;
        }
    }
}

ll mypow(ll a,ll b)
{
    ll ans=1;
    while(b)
    {
        if(b&1) ans=ans*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    return ans;
}

int main(void)
{
    int tt;
    scanf("%d",&tt);
    Prime();
    while(tt--)
    {
        int n,c;
        scanf("%d%d",&n,&c);
        printf("%lld\n",mypow(c,sum[n]));
    }
    return 0;
}