问题描述

I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

输入描述

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.

输出描述

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.

样例输入

2

1 2

112233445566778899 998877665544332211

样例输出

Case 1:

1 + 2 = 3

 

Case 2:

112233445566778899 + 998877665544332211 = 1111111111111111110

思路

简单的大数问题

代码

1 #include
     
       2 #include
      
        3  4 int main() 5 { 6     int n, j; 7     scanf("%d", &n); 8     for(j = 1; j <= n; j++)  9     {10         char ch1[1001], ch2[1001];11         scanf("%s", ch1);12         int s1 = strlen(ch1), i;13         int a[1001] = {
   0}, b[1001] = {
   0};14         for(i = s1-1; i >= 0; i--) { a[s1-i] = ch1[i] - '0'; }15         scanf("%s", ch2);16         int s2 = strlen(ch2);17         for(i = s2-1; i >= 0; i--) { b[s2-i] = ch2[i] - '0'; }18         printf("Case %d:\n%s + %s = ", j, ch1, ch2);19         int s = (s1>s2?s1:s2), temp=0;20         for(i = 1 ; i <= s; i++)21         {22             a[i] += b[i] + temp;23             temp = a[i] / 10;24             a[i] %= 10;25         }26         if(temp != 0) printf("%d",temp);27         for(i=s;i>0;i--) printf("%d",a[i]);28         if(j < n) printf("\n");29         printf("\n");30     }31 }