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: ", 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(" "); 58 } 59 return 0; 60 }