0x03大数问题（JAVA解决棋盘覆盖，A+B Problem II）

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常见的有棋盘覆盖和A+B问题，这类问题牵扯到的数值都比较大，如果用一般的数值类型，肯定输出不了，所以就要想一个办法，怎么把大数转换一下输出。

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A+B Problem II
时间限制：3000 ms | 内存限制：65535 KB
难度：3
描述
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
A,B must be positive.
输入
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.
输出
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation.
样例输入

2
1 2
112233445566778899 998877665544332211

样例输出

Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

偷懒的我，用java很简单

import java.util.*;
import java.math.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc=new Scanner (System.in);
        int l=sc.nextInt();
        for(int i=1;i<=l;i++){
            if(i!=1) System.out.println();
            BigInteger a,b;
            a=sc.nextBigInteger();
            b=sc.nextBigInteger();
            System.out.println("Case "+i+":");
            System.out.println(a+" + "+b+" = "+a.add(b));
        }
    }
}

java.math提供用于执行任意精度整数算法 (BigInteger) 和任意精度小数算法 (BigDecimal) 的类。

棋盘覆盖
时间限制：3000 ms | 内存限制：65535 KB
难度：3
描述
在一个2k×2k（1<=k<=100）的棋盘中恰有一方格被覆盖，如图1（k=2时），现用一缺角的2×2方格（图2为其中缺右下角的一个），去覆盖2k×2k未被覆盖过的方格，求需要类似图2方格总的个数s。如k=1时，s=1;k=2时，s=5
输入
第一行m表示有m组测试数据；
每一组测试数据的第一行有一个整数数k;
输出
输出所需个数s;
样例输入

3
1
2
3

样例输出

1
5
21

import java.util.*;
import java.math.*;
public class Main {
        public static void main(String[] args) {
            Scanner sc=new Scanner (System.in);
            int l=sc.nextInt();
            BigInteger m=BigInteger.valueOf(4);
            BigInteger n=BigInteger.valueOf(-1);
            BigInteger o=BigInteger.valueOf(3);
            for(int i=1; i<=l; i++) {
                if(i!=1) System.out.println();
                BigInteger s;
                
                int k=sc.nextInt();
                s=((m.pow(k)).add(n)).divide(o);
                System.out.println(s);
            }
        }
}

强大的java大数问题。