hdu4704之费马小定理+整数快速幂

 

Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 589    Accepted Submission(s): 292

Problem Description

 

 

Sample Input

2

 

 

Sample Output

2

Hint

1. For N = 2, S(1) = S(2) = 1. 2. The input file consists of multiple test cases.

 

 

/*分析:题目要求s1+s2+s3+...+sn;//si表示n划分i个数的n的划分的个数,如n=4,则s1=1,s2=3

假设An=s1+s2+s3+...+sn;

对于n可以先划分第一个数为n,n-1,n-2,...,1，则容易得出An=A0+A1+A2+A3+...+A(n-1);

=>A(n+1)=A0+A1+A2+A3+...+An =>An=2^(n-1);

由于n非常大,所以这里要用到费马小定理:a^(p-1)%p == 1%p == 1;//p为素数

所以2^n%m == ( 2^(n%(m-1))*2^(n/(m-1)*(m-1)) )%m == (2^(n%(m-1)))%m * ((2^k)^(m-1))%m == (2^(n%(m-1)))%m;//k=n/(m-1)

*/

#include<iostream>

#include<cstdio>

#include<cstdlib>

#include<cstring>

#include<string>

#include<queue>

#include<algorithm>

#include<map>

#include<iomanip>

#define INF 99999999

using namespace std;



const int MAX=100000+10;

const int mod=1000000000+7;

char s[MAX];



__int64 MOD(char *a,int Mod){

	__int64 sum=0;

	for(int i=0;a[i] != '\0';++i){

		sum=(sum*10+a[i]-'0')%Mod;

	}

	return sum;

}



__int64 FastPow(__int64 a,__int64 k){

	k=(k+mod)%mod;

	__int64 sum=1;

	while(k){

		if(k&1)sum=sum*a%mod;

		a=a*a%mod;

		k>>=1;

	}

	return sum;

}



int main(){

	while(scanf("%s",s)!=EOF){

		__int64 n=MOD(s,mod-1)-1;

		printf("%I64d\n",FastPow(2,n));

	}

	return 0;

}