HDOJ 1002 A + B Problem II

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 354745    Accepted Submission(s): 68820

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 Output

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

 

#include<iostream>#include<cstring>#include<string>#include<math.h>using namespace std;void add(string a,string b){int na=a.length();string c="";char ch;int flag=0;int aa;int bb;int cc;int i;for(i=na-1;i>=0;i--){aa=a[i]-'0';bb=b[i]-'0';cc=aa+bb+flag;c=char(cc%10+48)+c;if(cc>=10){flag=1;}else{flag=0;}}if(flag==1){c="1"+c;}cout<<c;cout<<endl; } int main(){void add(string a,string b);string a,b;int n;int na,nb;cin>>n;int num=1;int i;for(i=0;i<n;i++){cin>>a>>b;string ta=a;string tb=b;na=a.length();nb=b.length();if(na>nb){for(int j=0;j<na-nb;j++){b='0'+b; } }if(na<nb){for(int j=0;j<nb-na;j++){a='0'+a;}}cout<<"Case "<<num<<":\n";cout<<ta<<" + "<<tb<<" = ";add(a,b);num++;if(i!=n-1) cout<<endl;}return 0; }