URAL 1528 Sequence

SequenceCrawling in process... Crawling failed Time Limit:3000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice URAL 1528

Description

You are given a recurrent formula for a sequence f:

f(n) = 1 + f(1)g(1) + f(2)g(2) + … + f(n−1)g(n−1),

where g is also a recurrent sequence given by formula

g(n) = 1 + 2g(1) + 2g(2) + 2g(3) + … + 2g(n−1) − g(n−1)g(n−1).

It is known that f(1) = 1, g(1) = 1. Your task is to find f(n) mod p.

Input

The input consists of several cases. Each case contains two numbers on a single line. These numbers are n (1 ≤ n ≤ 10000) and p (2 ≤ p ≤ 2·10⁹). The input is terminated by the case with n = p = 0 which should not be processed. The number of cases in the input does not exceed 5000.

Output

Output for each case the answer to the task on a separate line.

Sample Input

input	output
1 2 2 11 0 0	1 2

题意：如题。

思路：哇。

一開始看错题啦。

不难发现f（n）的通项啦。

AC代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <stdlib.h>

using namespace std;

int main(){
    int n,p;
    while(~scanf("%d%d",&n,&p)){
        if(n==0&&p==0) break;
        long long ans=1;
        for(int i=2;i<=n;i++){
            ans*=i%p;
            ans%=p;
        }
        printf("%d\n",ans%p);
    }
    return 0;
}