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;
}
}