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;
	}
}