bzoj1005: [HNOI2008]明明的烦恼

题目
题解
calc(x,y)表示 y ! 中含有因数x的个数

#include<bits/stdc++.h>
using namespace std;
int n,x,i,s,t,sum,cnt,pri[1002],b[1002],d[1002],k,j,temp;
struct NUM{
    int a[100002],t;
}ans,A,T;
void cheng(NUM &x,int y){
    int i;
    for (i=0;i<x.t;i++) x.a[i]*=y;
    for (i=0;i<x.t || x.a[i];i++) x.a[i+1]+=x.a[i]/10,x.a[i]%=10;
    x.t=i;
}
int calc(int x,int y){
    int s=0;
    for (;y>=x;y/=x,s+=y);
    return s;
}
void print(NUM x){
    for (int i=x.t-1;i>=0;i--) putchar(x.a[i]|48);
    putchar('\n');
}
int main(){
    scanf("%d",&n);
    if (n==1){
        scanf("%d",&x);
        puts(x<1?"1":"0");
        return 0;
    }
    for (i=1;i<=n;i++){
        scanf("%d",&x);
        if (!x){
            puts("0");
            return 0;
        }
        if (x!=-1) d[++cnt]=x-1,sum+=d[cnt];
    }
    if (n-2<sum){
        puts("0");
        return 0;
    }
    ans.a[0]=ans.t=1;
    for (i=2;i<=n;i++){
        if (!b[i]){
            pri[k++]=i;
            s=calc(i,n-2)-calc(i,n-2-sum);
            for (j=1;j<=cnt;j++) s-=calc(i,d[j]);
            while (s--) cheng(ans,i);
        }
        for (j=0;j<k && i*pri[j]<=n;j++){
            b[i*pri[j]]=1;
            if (i%pri[j]==0) break;
        }
    }
    for (i=0;i<n-2-sum;i++) cheng(ans,n-cnt);
    print(ans);
}

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转载自blog.csdn.net/xumingyang0/article/details/80821748