URAL 1528 Sequence

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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 + 2 g(1) + 2 g(2) + 2 g(3) + … + 2 g( 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 22 110 0	12

題意：如題。

思路：哇。一開始看錯題啦。不難發現f（n）的通項啦。

AC代碼：

#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;}