HDU 4704 Sum （费马定理+快速幂）

Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 647    Accepted Submission(s): 320

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.

 

 

Source

2013 Multi-University Training Contest 10

 

 

思路：一道整数划分题目，不难推出公式：2^(n-1)，

根据费马小定理：(2,MOD)互质，则2^(p-1)%p=1,于是我们可以转化为：2^(n-1)%MOD=2^((n-1)%(MOD-1))%MOD,从而用快速幂求解。

 公式2^(n-1) % MOD;

可先对(n-1)%(MOD-1)

 

import java.io.*;

import java.util.*;

import java.math.*;

public class Main {

	BigInteger n;

	String s="";

	BigInteger one=BigInteger.valueOf(1);

	BigInteger Mod=BigInteger.valueOf((long)(1e9+7));

	BigInteger Mod1=BigInteger.valueOf((long)(1e9+6));

	public static void main(String[] args) {

		new Main().work();

	}

	void work(){

		Scanner sc=new Scanner(new BufferedInputStream(System.in));

		while(sc.hasNext()){

			s=sc.next();

			n=BigInteger.valueOf(0);

			for(int i=0;i<s.length();i++){

				n=(n.multiply(BigInteger.valueOf(10)).add(BigInteger.valueOf(s.charAt(i)-'0'))).mod(Mod1);

			}

			long num=n.longValue()-1;

			System.out.println(pow(BigInteger.valueOf(2),num).mod(Mod)); 

		}

	}

	BigInteger pow(BigInteger a,long b){

		BigInteger sum=BigInteger.ONE;

		while(b!=00){

			if((b&1)!=0){

				sum=sum.multiply(a).mod(Mod);

			}

			a=a.multiply(a).mod(Mod);

			b>>=1;

		}

		return sum;

	}

}