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 ;

}