CSP-S Analog 61

They explode

simulation

Thank detailed solution to a problem rvalue seniors

　　The problem is that the first non-low from the start 0 output, that is, n-$! $ 10 is removed (removing the suffix 0), i.e. all the Ext $ factors (10, n-!) $

　　And the number of outputs in the k == 1 when a $ Ext (10, n!) \% 10 $, k == 2 when a $ Ext (10, n!) \% 100 $, k == 3 when ......

　　10, for example

　　The $ \% $ 10 can first find $ \% 2 $ and $ \ $ 5% of the answers, the final answer with CRT demand

　　$ Ext (10, n!) \% 2 $ a little bit special judge (usually 0, but may not be small, so small you can run violence)

　　$ Ext (10, n!) \% 5 $ you need to change the formula it should

　　　　　　　　　$$\begin{array}{rl}Ext(10,n!)&= \frac{n!}{10^{FP(10,n!)}}&= \frac{n!}{2^{FP(10,n!)}5^{FP(10,n!))}}&= \frac{n!}{2^{FP(5,n!)}5^{FP(5,n!)}}&=\frac{Ext(5,n!))}{2^{FP(5,n!)}}\end{array}$$

　　$ FP (5, n!) $ Can be determined in time $ log_5 ^ n $, then the $ 2 ^ {FP (5, n!)} $ The inverse element there, leaving only $ Ext (5, n !) $

　　$ Ext (k, n) $ the nature $ Ext (ab, n) = Ext (a, n) * Ext (b, n) (a, b is a prime number) $

　　所以$$ \begin{array}{rl} Ext(5,n!)&= \prod\limits_{k=1}^{n} Ext(5,k) \\&=\prod\limits_{k\in[1,n],5|k} Ext(5,k)\times \prod\limits_{k\in[1,n],5\nmid k}Ext(5,k)\\&=Ext(5,(\frac{n}{5})!)\times \prod\limits_{k\in[1,n],5\nmid k}Ext(5,k)\\&=Ext(5,(\frac{n}{5})!)\times \prod\limits_{k\in[1,n],5\nmid k} k\end{array} $$

　　and

　　所以$$Ext(5,n!)= Ext(5,(\frac{n}{5})!) \times (\prod \limits_{k\in[1,5],5\nmid k} k)^{\lfloor \frac{n}{5} \rfloor} \times \prod \limits_{k\in[1,n\%5],5\nmid k} k (mod 5) $$

　　 Find out the final $ Ext (5, n!) $ Multiplying the front seeking $ 2 ^ {FP (5, n!)} $ Is the inverse element thus obtained $ Ext (10, n!) \% 5 $

　　And $ Ext (10, n!) \% 2 $ ran CRT (big God do not really CRT) obtained $ Ext (10, n!) \% Solution of 10 $

　　k == 2 and k == 3 Similarly

　　Then to k == 3 Case Study

　　$$Ext(5,n!)= Ext(5,(\frac{n}{5})!) \times (\prod \limits_{k\in[1,125],5\nmid k} k)^{\lfloor \frac{n}{125} \rfloor} \times \prod \limits_{k\in[1,n\%125],5\nmid k} k (mod 125) $$

　　The last Bo Fuli (Code)

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int T,k,a[110],FP[110],b[110],fac[150];
long long ans[11000];
int c[5],p[5],phi;
inline void read(){
    register char r;
    while(r=getchar(),r<'0'||r>'9');
    a[0]=0;
    a[++a[0]]=r^48;
    while(r=getchar(),r>='0'&&r<='9') a[++a[0]]=r^48;
}
long long qpow(long long a,long long b,long long mod){
    long long ans=1;
    while(b){
        if(b&1) ans=ans*a%mod;
        b>>=1;
        a=a*a%mod;
    }
    return ans;
}
int div(int a[],int b){
    int tmp=0;
    for(register int i=a[0];i;i--){
        tmp=tmp*10+a[i];
        a[i]=tmp/b,tmp%=b;
    }
    while(a[a[0]]==0&&a[0]>=1) a[0]--;
    return tmp;
}
void jia(int a[],int b[]){
    int tmp=0;
    for(register int i=1;i<=max(a[0],b[0]);i++){
        tmp=a[i]+b[i]+tmp;
        b[i]=tmp%10;
        tmp/=10;
    }
    b[0]=max(a[0],b[0]);
    while(tmp) b[++b[0]]=tmp%10,tmp/=10;
}
int MOD(int a[],int mod){
    return div(a,mod);
}
int work_FP(){
    memset(FP,0,sizeof(FP));
    memcpy(b,a,sizeof(b));
    while(b[0]){
        div(b,5);
        jia(b,FP);
    }
    return MOD(FP,phi);
}
void work_to_2(){
    if(k==1) p[1]=2,p[2]=5,phi=4;
    else if(k==2) p[1]=4,p[2]=25,phi=20;
    else if(k==3) p[1]=8,p[2]=125,phi=100;
    if(a[0]>1) c[1]=0;
    else{
        c[1]=1;
        for(register int i=1;i<=a[1];i++) c[1]=c[1]*i%p[1];
    }
}
void work_to_5(){
    int w=work_FP()+phi;
    w=qpow(qpow(2,w,p[2]),phi-1,p[2]);
    fac[0]=1;
    for(register int i=1;i<=p[2];i++){
        fac[i]=fac[i-1];
        if(i%5!=0) fac[i]=fac[i]*i%p[2];
    }
    
    c[2]=1;
    while(a[0]){
        memcpy(b,a,sizeof(b));
        memset(FP,0,sizeof(FP));
        div(a,5);
        c[2]=c[2]*fac[div(b,p[2])]%p[2];
        jia(b,FP);
        c[2]=c[2]*qpow(fac[p[2]],MOD(FP,phi),p[2])%p[2];
    }
    c[2]=c[2]*w%p[2];
}
void work(){
    for(register int i=a[0];i>=1;i--) b[++b[0]]=a[i];
    memcpy(a,b,sizeof(a));
    work_to_2();
    work_to_5();
}
void exgcd(int a,int b,int &x,int &y){
    if(!b){
        x=1;
        y=0;
        return;
    }
    exgcd(b,a%b,x,y);
    int t=y;
    y=x-a/b*y;
    x=t;
    return;
}
int CRT(){
    int sum=p[1]*p[2];
    int ans=0;
    for(register int i=1;i<=2;i++){
        int a=sum/p[i],x,y;
        exgcd(a,p[i],x,y);
        ans=(ans+a*x*c[i]%sum)%sum;
    }
    return (ans%sum+sum)%sum;
}
void out(int x){
    if(k==1) printf("%d\n",x%10);
    if(k==2) printf("%d%d\n",(x/10)%10,x%10);
    if(k==3) printf("%d%d%d\n",(x/100)%10,(x/10)%10,x%10);
}
int main(){
    //freopen("3.out","w",stdout);
    scanf("%d",&T);
    ans[0]=1;
    for(register int i=1;i<=10000;i++){
        ans[i]=ans[i-1]*i;
        while(ans[i]%10==0) ans[i]/=10;
        ans[i]%=((int)1e8);
    }
    while(T--){
        read();
        scanf("%d",&k);
        if(a[0]<=4){
            int n=0;
            for(register int i=1;i<=a[0];i++) n=n*10+a[i];
            out(ans[n]); 
        }
        else{
            work();
            out(CRT());
        }
    }
}

View Code

T3 甜圈

　　要求完成的任务必须按顺序而且只能做一次，用hash表示状态

　　将甜圈编号作为线段树下标，经典的线段树操作就可以了

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