A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 399645 Accepted Submission(s): 77352
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
用JAVA
import java.math.BigInteger; import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner in = new Scanner(System.in); int n; BigInteger a,b; n=in.nextInt(); int i=0; while(n>0) { n--; a=in.nextBigInteger(); b=in.nextBigInteger(); i++; System.out.println("Case" + " " + i + ":"); if(n>0) { System.out.println(a + " + " + b + " = "+a.add(b)); System.out.println(); } else { System.out.println(a + " + " + b + " = "+a.add(b)); } } } }
C++:
要审清题意，注意输出格式！ #include<string> #include <cstdio> #include<iostream> using namespace std; string add(string a,string b) { int len1=a.length(); int len2=b.length(); if(len1>len2) { for(int i=1;i<=len1-len2;i++) b="0"+b; } else { for(int i=1;i<=len2-len1;i++) a="0"+a; } int len=a.length(); int cf=0,t; string str; for(int i=len-1;i>=0;i--) { t=a[i]-'0'+b[i]-'0'+cf; cf=t/10; t%=10; str=char(t+'0')+str; } if(cf!=0) str=char(cf+'0')+str; return str; } int main() { int t; cin>>t; int k=0; while(t--) { k++; string a,b; cin>>a>>b; string str; str=add(a,b); cout<<"Case "<<k<<":"<<endl; cout<<a<<" + "<<b<<" = "<<str<<endl; if(t) cout<<endl; } return 0; }
转载于:https://www.cnblogs.com/Weixu-Liu/p/9165365.html
