A + B Problem II
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
1 #include <iostream> 2 #include <cstring> 3 using namespace std; 4 int main() 5 { 6 int len1,len2,p; 7 int i,j,t; 8 char a[1000],b[1000],c[1001]; 9 cin>>t; 10 for(j=1;j<=t;j++) 11 { 12 cin>>a>>b; 13 cout<<"Case"<<" "<<j<<":"<<endl; 14 cout<<a<<" "<<"+"<<" "<<b<<" "<<"="<<" "; 15 len1=strlen(a)-1; 16 len2=strlen(b)-1; 17 p=0; 18 for(i=0;len1>=0||len2>=0;i++,len1--,len2--) 19 { 20 if(len1>=0&&len2>=0) 21 c[i]=a[len1]+b[len2]-'0'+p; 22 if(len1>=0&&len2<0) 23 c[i]=a[len1]+p; 24 if(len2>=0&&len1<0) 25 c[i]=b[len2]+p; 26 p=0; 27 if(c[i]>'9') 28 { 29 c[i]=c[i]-10; 30 p=1; 31 } 32 } 33 if(p==1) 34 cout<<"1"; 35 while(i--) 36 cout<<c[i]; 37 if(j<t) 38 cout<<endl<<endl; 39 else 40 cout<<endl; 41 } 42 return 0; 43 }