~~~
# A + B Problem II
Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 273307????Accepted Submission(s): 52784
~~~
Problem Description
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
Input
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.
Output
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. Output a blank line between two test cases.
Sample Input
~~~
2
1 2
112233445566778899 998877665544332211
~~~
Sample Output
~~~
Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
~~~
~~~
import java.math.BigDecimal;
import java.util.Scanner;
// 代碼提交時改為Main且去掉當前文件的包名
public class Main {
public static void main(String[] args) {
/**
* 從題目中可以看出來,數據比較大,只是用普通的整形或者別的恐怕是無能為力,不過BigDecimal對于數據較大的數處理十分的方便
* 而且封裝好了函數只需要調用就可以了
*/
Scanner scanner = new Scanner(System.in);
int num = scanner.nextInt();
// 這里的二維數組也需要有些講究,一般一行中查找不同的列的速度要比一列中查找不同的行的數據要快
BigDecimal[][] bigNum = new BigDecimal[num][2];
for(int i = 0; i < num; i++)
{
bigNum[i][0] = scanner.nextBigDecimal();
bigNum[i][1] = scanner.nextBigDecimal();
}
// 輸出的時候注意空格,以及的空行
for(int j = 0; j < num; j++)
{
if(j != 0)
System.out.println();
System.out.println("Case "+(j+1)+":");
System.out.println(bigNum[j][0] +" + "+ bigNum[j][1]+" = "+ bigNum[j][0].add(bigNum[j][1]));
}
}
}
~~~
- 前言
- 求和的問題ACM
- A+B問題acm
- 1091ACM求和
- 杭電ACM1092求和問題詳解
- ACM杭電的1093求和問題
- 杭電ACM1094計算A+B的問題
- 杭電ACM1095解決A+B問題
- 杭電ACM1096求和問題
- 杭電Acm1001解決求和的問題
- 杭電ACM1008電梯問題C++
- 杭電ACM大賽2000關于ASCII碼排序的問題
- 杭電ACM2006奇數的乘積
- 杭電ACM數值統計2008
- 杭電ACM1019求最大公約數
- 杭電ACM1108求最小公倍數
- 杭電ACM2035人見人愛的A^B
- 杭電ACM1061N^N求最右邊的數的問題
- 杭電ACM1021裴波納挈數AGAIN
- 杭電ACm1005求f(n)非遞歸
- 杭電ACM1071The area---------求積分面積
- 杭電ACM吃糖果問題
- 杭電ACm求數列的和2009
- 杭電ACM多項式求和--》2011
- 杭電ACM。。。sort
- 杭電ACM1004
- 杭電ACM2043密碼的問題已經AC
- 杭電ACM2041樓梯問題
- 動態規劃C++::杭電ACM1003
- 杭電ACM----2018母牛的故事
- 杭電ACM2007平方和與立方和
- 盧卡斯隊列
- 全國軟件2. 三人年齡
- 全國軟件3. 考察團組成
- 全國軟件--微生物增殖
- 全國軟件填寫算式
- 全國軟件-----------猜生日
- 全國軟件---------歐拉與雞蛋
- Java經典算法四十例編程詳解+程序實例
- 杭電ACMA + B Problem II問題解析
- 杭電ACM1018BigNumber解析
- 杭電ACM1088 Write a simple HTML Browser Java
- 杭電ACM1106排序Java代碼
- 杭電ACM 1012 u Calculate e java
- 杭電ACM 1020 Encoding java解析
- 杭電1047 An Easy Task - java 解讀
- 杭電ACM 1040 As Easy As A+B java 解讀
- 杭電ACM 1041 Computer Transformation java代碼詳解AC
- 杭電ACM 1030 Delta-wave java代碼解析