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1002 A + B Problem II

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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;
}

参考

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