HeHe HDU - 2879（积性）

最新推荐文章于 2020-01-13 09:21:46 发布

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本文介绍了一种通过模运算解决特定方程X^2≡X(modN)的解决方案计数问题的方法，并提供了一个高效的算法实现。该算法首先通过筛选获取质数，然后利用快速幂运算计算最终结果。

In the equation X^2≡X(mod N) where x∈[0,N-1], we define He[N] as the number of solutions.
And furthermore, define HeHe[N]=He[1]*……*He[N]
Now here is the problem, write a program, output HeHe[N] modulo M for a given pair N, M.
Input
First line: an integer t, representing t test cases.
Each test case contains two numbers N (1<=N<=10^7) and M (0<M<=10^9) separated by a space.
Output
For each test case, output one line, including one integer: HeHe[N] mod m.
Sample Input
1
2 3
Sample Output
2

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<bitset>
#include<vector>
#define N 10000500
using namespace std;
typedef long long  ll;
ll fac[100];
int k=0;
ll n,m;
vector<int>prime;
bitset<N> vis;
void getPrime()
{
    for(int i=2;i<N;i++)
        if(!vis[i])
        {
            prime.push_back(i);
            for(int j=i+i;j<N;j+=i)
                vis[j]=true;
        }
}
ll P(ll a,ll b,ll m)
{
    ll ans=1;
    while(b)
    {
        if(b&1)
            ans=ans*a%m;
        a=a*a%m;
        b>>=1;
    }
    return ans;
}
ll work(ll x)
{
    ll b=0;
    for(int i=0;prime[i]<=x;i++)
        b+=x/prime[i];

    return P(2,b,m);
}
int main()
{
    int t;
    getPrime();
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld%lld",&n,&m);
        //cout<<"hah";
        printf("%lld\n",work(n));
    }
    return 0;
}