每日一練ACM 2019.04.13
阿新 • • 發佈:2019-04-15
import RoCE scrip 計算 表示 個數 there test 超過 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
2019.04.13
第1002題:A+B Proble Ⅱ
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.
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
1 2
112233445566778899 998877665544332211
Sample Output
Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
題目解析:
輸入的第一行:T為實例的數量,再輸入T行的例子,每行有2個正整數(他們很大,所以不能用32位的整數來表示),要求計算兩個數的和並輸出,保證輸入的每個整數的長度不超過1000。
解:
package Acm; import java.math.BigInteger; import java.util.Scanner;public class test3 { public static void main(String[] args) { // TODO Auto-generated method stub Scanner input=new Scanner(System.in); int n=input.nextInt(); for(int i=0;i<n;i++) { BigInteger a=input.nextBigInteger(); BigInteger b=input.nextBigInteger(); if(i==n-i) { System.out.print("case"+(i+1)+"\r\n"+a+"+"+b+"="+a.add(b)); } else { System.out.println("case"+(i+1)+"\r\n"+a+"+"+b+"="+a.add(b)); } } } }
每日一練ACM 2019.04.13