BZOJ 2186 [Sdoi2008]沙拉公主的困惑 线性逆元

题意:链接

方法:线性筛逆元

解析:

SB卡常数题

不想多说什么正常人都会做

ans=ϕ(m!)∗n!m!

ans=m!∗n!m!∗∏Pi−1Pi(Pi|m!)

ans=n!∗∏Pi−1Pi(Pi|m!)

ans=n!∗∏((Pi−1)∗inv[Pi])

代码:

#include<cstdio>#include<cmath>#include<ctime>#include<cstring>#include<iostream>#include<algorithm>#define inf 1000000000#define N 10000000#define ll long longusing namespace std;int read(){    int x=0;char ch=getchar();    while(ch<'0'||ch>'9')ch=getchar();    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x;}int T,R,n,m,cnt;int fac[10000005],ine[10000005],pri[500005],ans[10000005];bool mark[10000005];void exgcd(int a,int b,int &x,int &y){    if(b==0){x=1;y=0;return;}    exgcd(b,a%b,x,y);    int t=x;x=y;y=t-a/b*y;}int getine(int t){    int x,y;    exgcd(t,R,x,y);    return (x%R+R)%R;}void pre(){    fac[1]=1;for(int i=2;i<=N;i++)fac[i]=(ll)fac[i-1]*i%R;    ine[1]=1;    for(int i=2;i<=N;i++)    {        if(!mark[i])pri[++cnt]=i,ine[i]=getine(i);        for(int j=1;pri[j]*i<=N&&j<=cnt;j++)        {            mark[pri[j]*i]=1;            //ine[pri[j]*i]=(ll)ine[pri[j]]*ine[i]%R;            if(i%pri[j]==0)break;        }    }    ans[1]=1;    for(int i=2;i<=N;i++)    {        ans[i]=ans[i-1];        if(!mark[i])ans[i]=(ll)ans[i]*(i-1)%R*ine[i]%R;    }}int main(){    T=read();R=read();    pre();    while(T--)    {        n=read();m=read();        printf("%d\n",(ll)fac[n]*ans[m]%R);    }    return 0;}