A+B Problem Plus [Medium]

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.

解答要求
时间限制：1000ms, 内存限制：100MB
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. Output a blank line between two test cases.

样例
输入样例 1 复制

2
1 2
112233445566778899 998877665544332211
输出样例 1

Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

反向打印与补零

这里在原数组正着一位一位的加上，同时注意超过的位数补全为0，之后输出时反向打印出来。

 #define MAX_NUM 100
int BigNumberAdd(char *a, char *b, int *calcResult)
{
int lenA = strlen(a);
int lenB = strlen(b);
int carry = 0;
int resLen = 0;
while (lenA > 0 || lenB > 0 || carry > 0) {
    int chA = lenA > 0 ? a[lenA - 1] - '0' : 0;
    int chB = lenB > 0 ? b[lenB - 1] - '0' : 0;
    int temp = chA + chB + carry;        
    carry = temp / 10;
    calcResult[resLen++] = temp % 10;
    lenA--;
    lenB--;
}
return resLen;
}
int main()
{
int t;
if (scanf_s("%d", &t) == EOF) {
    return 0;
}
int count = 0;
while (t--) {
    char a[MAX_NUM];
    char b[MAX_NUM];
    if (scanf_s("%s%s", a, sizeof(a), b, sizeof(b)) == EOF) {
        continue;
    }
    int calcResult[MAX_NUM];
    int resLen = BigNumberAdd(a, b, calcResult);
    count++;
    printf("Case %d:\n", count);
    printf("%s + %s = ", a, b);
    for (int i = resLen - 1; i >= 0; i--) {
        printf("%d", calcResult[i]);
    }
    printf("\n\n");
}
return 0;

}