题目描述

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

A,B must be positive.

输入

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

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.

样例输入

21 2112233445566778899 998877665544332211

样例输出

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

1 /*对于此题做法有两种： 2  其一，使2字符串的中的字符数字减去'0'，逐个相加大于等于10的可以使本位减10，下一位自增1，后面的处理就非常简单了； 3  其二，便是读入字符串后先让各个字符减'0'，一一对应存入整形数组中;之后再相加。 4     对于2种方法大致是相同的，都要从后面向前加，逢十进位，以及数组初始化均要初始为0，一边方便运算。 5  下面用法一，代码如下：*/ 6 #include
      
        7 #include
       
         8  9 int main( ) {10     int n, i,len1, len2, j, k, pi, t;11     char str1[1010], str2[1010];12     scanf("%d", &n);13     for(i = 1; i <= n; i++) {14         int a[1200] = {
    0};15         int flag = 0;16         scanf("%s%s", str1, str2);//以字符串形式读入17         printf("Case %d:\n", i);18         len1 = strlen(str1);19         len2 = strlen(str2);20         printf("%s + %s = ", str1, str2);21         j = len1-1;22         k = len2-1;23         pi = 0;24         while(j >= 0 && k >= 0) { //开始相加25             if(a[pi] + (str1[j]-'0') + (str2[j]-'0') >= 10) { //相加后大于10的26                 a[pi] = a[pi] + (str1[j] -'0')+(str2[k] - '0')-10;27                 a[pi+1]++;28             }29             else30                 a[pi] = a[pi] + (str1[j] - '0')+(str2[k]-'0');31             pi++; k--; j--;32         }33         if(j>=0) {34             for(t = j; t >= 0; t--) {35                 a[pi] = a[pi]+(str1[t]-'0');36                 pi ++;37             }38         }39         else if(k >= 0) {40             for(t = k; t >= 0; t--) {41                 a[pi] = a[pi] + str2[t]-'0';42                 pi++;43             }44         }45         else if(a[pi]!=0)//对于位数相同2个数加后最高位大于10的46             pi++;47         for(t = pi - 1; t >= 0; t--) {48             if(a[t]==0&&flag==0)49                 continue;50             else51             {52                 flag=1;53                 printf("%d", a[t]);54             }55         }56         if(i!=n)//对于2组之间加空行的情况57             printf("\n");58     }59     return 0;60 }