http://bailian.openjudge.cn/PRactice/4010

題意：已知長度最大為200位的正整數n，請求出2011^n的后四位。

第一次用java BigInteger類，參考了一下BigInteger的一些方法，就可以寫出大數快速冪。

import java.util.Scanner;import java.math.BigInteger;public class Main {    static BigInteger base = new BigInteger("2011");    static BigInteger MOD = new BigInteger("10000");    static BigInteger zero = new BigInteger("0");    static BigInteger one = new BigInteger("1");    static BigInteger two = new BigInteger("2");    public static BigInteger FastMul(BigInteger n){        if(n.equals(one)){            return base;        }        if(n.mod(two).equals(zero)){            BigInteger ret = FastMul(n.divide(two));            return ret.multiply(ret).mod(MOD);        }        else{            return base.multiply(FastMul(n.subtract(one))).mod(MOD);        }    }    public static void main(String[] args) {        Scanner sc = new Scanner(System.in);        int k = sc.nextInt();        for(int i=0; i<k; ++i){            BigInteger n = sc.nextBigInteger();            BigInteger ans = FastMul(n);            System.out.println(ans.mod(MOD));        }    }}
ss