为什么会是WA啊??????
/****************************************************************************************Problem DescriptionI have a very simple problem for you. Given two integers A and B, your job is tocalculate the Sum of A + B.InputThe first line of the input contains an integer T(1<=T<=20) which means the numberof 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 processthem by using 32-bit integer. You may assume the length of each integer will notexceed 1000.OutputFor 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 ablank line between two test cases.Sample Input21 2112233445566778899 998877665544332211Sample OutputCase 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110****************************************************************************************/#include <stdio.h>#include <string.h>int main (void){ char a[1000], b[1000], c[1001]; // 最高位和可能会进位, 所以要多一字节存进位. int n, i, j, k; int alength, blength, clength; char carry = 0; // 进位标识, 只可能是1/0 while(scanf("%d", &n) != EOF) { for (i = 0; i < n; i++) // 处理 n 组数 { memset(a, 0, sizeof(a)); // 将所有分配的内存置 0. memset(b, 0, sizeof(b)); memset(c, 0, sizeof(c)); scanf("%s %s", a, b); // 输入第 1 和 2 两数. printf("Case %d:\n", i + 1); alength = strlen(a); // 获取数字长度 blength = strlen(b); clength = (alength >= blength) ? alength : blength; for (j = alength - 1, k = blength - 1; j >= 0 && k >= 0; j--, k--) { if (a[j] + b[k] - '0' + carry > '9') { c[clength--] = (a[j] - '0' + b[k] - '0' + carry) % 10 + '0'; carry = 1; } else { c[clength--] = (a[j] - '0' + b[k] - '0' + carry) % 10 + '0'; carry = 0; } } while (j >= 0 && clength >= 0) { // 当 a[] 比 b[] 长时 if (a[j] + carry > '9') { c[clength--] = (a[j--] + carry - '0') % 10 + '0'; carry = 1; } else { c[clength--] = a[j--] + carry; carry = 0; } } while (k >= 0 && clength >= 0) { // 当 b[] 比 a[] 长时 if (a[k] + carry > '9') { c[clength--] = (b[k--] + carry - '0') % 10 + '0'; carry = 1; } else { c[clength--] = b[k--] + carry; carry = 0; } } c[clength] = carry + '0'; k = 0; // 就不多定义变量了, 复用吧. while (c[k] == '0') { // 去除结果中的前导 0. k++; } printf("%s + %s = %s\n\n", a, b, c + k); carry = 0; // 处理下组数之前清零进位位 } } return 0;}