问题 C: A+B Problem II

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

2
1 2
112233445566778899 998877665544332211

样例输出

Case 1:
1 + 2 = 3
Case 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<stdio.h>
 7 #include<string.h>
 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             else
30                 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             else
51             {
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 }