# 1002 A + B Problem II

1002 A + B Problem 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.

## 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
``````

## 代码

``````#include
using
namespace std;

int
main()
{
int k, n, l1, l2, l;
int n1[1002], n2[1002];
char str1[1002], str2[1002];
cin>>n;
k=n;
while(k--)
{
memset(n1, 0, sizeof(n1));
memset(n2, 0, sizeof(n2));
cin>>str1;
cin>>str2;
l1=strlen(str1);
l2=strlen(str2);
if(l1>l2)
l=l1;
else
l=l2;
for(int i=l1-1, j=0; i>=0; i--, j++)
n1[j]=str1[i]-'0';
for(i=l2-1, j=0; i>=0; i--, j++)
n2[j]=str2[i]-'0';
for(i=0; i
{
n1[i]+=n2[i];
if(n1[i]>9)
{
n1[i+1]+=n1[i]/10;
n1[i]=n1[i];
}
}

n1[l-1]=n1[l-1]+n2[l-1];
if(n1[l-1]>9)
{
l++;
n1[l-1]+=n1[l-2]/10;
n1[l-2]=n1[l-2];
}
char s[1002];
for(i=l-1, j=0; i>=0; i--, j++)
s[j]=n1[i]+'0';
s[l]='\0';
cout<<"Case "<<n-k<<":"<<endl;
cout<<str1<<" + "<<str2<<" ="<<s<<endl;
if(k!=0)
cout<<endl;
}
return 0;
}
``````
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