[SDOI2008]沙拉公主的困惑 线性筛_欧拉函数_逆元_快速幂

Code:

#include<cstdio>
using namespace std;
typedef long long ll;
const int maxn=10000000+1;
long long mod;
ll   fac[maxn];
ll  inv[maxn];
ll  anss[maxn];
int cnt,prime[maxn];
bool vis[maxn];
ll pow(ll base,ll k)
{
    ll ans=1;
    while(k)
    {
        if(k&1)
            ans=(ll)(ans*base)%mod;
        k/=2;
        base=(ll)(base*base)%mod;
    }
    return ans;
}
ll get_inv(ll a){return pow(a,mod-2);}
void init()
{
    fac[1]=1;
    inv[1]=1;
    for(int i=2;i<maxn;++i)
    {
        fac[i]=(fac[i-1]*i)%mod;
        if(!vis[i])prime[++cnt]=i,inv[i]=get_inv(i);
        for(int j=1;j<=cnt&&prime[j]*i<maxn;++j)
        {
            vis[prime[j]*i]=1;
            if(i%prime[j]==0)break;
        }
    }
    anss[1]=1;
    for(int i=2;i<maxn;++i)                         //找M
    {
    	if(!vis[i])
    	{
    		anss[i]=(anss[i-1]*(i-1))%mod;
    		anss[i]=(anss[i]*inv[i])%mod;
    	}
    	else anss[i]=anss[i-1];
    }
}
int main()
{
    int T;
    scanf("%d %lld",&T,&mod);
    init();
    while(T--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        ll ans;
        ans=(anss[m]*fac[n])%mod;
        printf("%lld
",ans);
    }
    return 0;
}
原文地址:https://www.cnblogs.com/guangheli/p/9845075.html