URAL 1528 Sequence

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

SubmitStatus 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 findf(n) mod p.

Input

The input consists of several cases. Each case contains two numbers on a single line. These numbers aren (1 ≤ n ≤ 10000) and p (2 ≤ p ≤ 2·10⁹). The input is terminated by the case withn = 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

inputoutput

1 22 110 0

12

题意：如题。

思路：哇。一开始看错题啦。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;}