BZOJ 2186 沙拉公主的困惑(预处理逆元+欧拉函数)

题意:求1-n!里与m!互质的数有多少？(m<=n<=1e6).

因为n!%m!=0,所以题目实际上求的是phi(m!)*n!/m!.

预处理出这些素数的逆元和阶乘的模即可。

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <bitset>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-
# define MOD
# define INF
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<,l,mid
# define rch p<<|,mid+,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
const int N=;
//Code begin...

int fac[N], pri[N], inv[N], ans[N];
bool vis[N];
int P, tot;
void init(int n)
{
    fac[]=;
    FOR(i,,n) fac[i]=(LL)fac[i-]*i%P;
    inv[]=;
    FOR(i,,n){
        if(!vis[i])pri[++tot]=i;
        for(int j=; pri[j]*i<=n&&j<=tot; ++j){
            vis[pri[j]*i]=;
            if(i%pri[j]==) break;
        }
    }
    for(int i=; i<=n&&i<P; ++i) inv[i]=(P-(LL)P/i*inv[P%i]%P);
    ans[]=;
    FOR(i,,n) {
        ans[i]=ans[i-];
        if(!vis[i])ans[i]=(LL)ans[i]*(i-)%P*inv[i%P]%P;
    }
}
int main()
{
    int T, n, m; scanf("%d%d",&T,&P); init(N-);
    while(T--){
        scanf("%d%d",&n,&m);
        printf("%d\n",(LL)fac[n]*ans[m]%P);
    }
    return ;
}