POJ 1504 Adding Reversed Numbers(내 수제의 길 - 역방향 수 고정밀 덧셈)

Adding Reversed Numbers
Time Limit: 1000MS
 
Memory Limit: 10000K
Total Submissions: 11366
 
Accepted: 6323
Description
The Antique Comedians of Malidinesia prefer comedies to tragedies. Unfortunately, most of the ancient plays are tragedies. Therefore the dramatic advisor of ACM has decided to transfigure some tragedies into comedies. Obviously, this work is very hard because the basic sense of the play must be kept intact, although all the things change to their opposites. For example the numbers: if any number appears in the tragedy, it must be converted to its reversed form before being accepted into the comedy play. 
Reversed number is a number written in arabic numerals but the order of digits is reversed. The first digit becomes last and vice versa. For example, if the main hero had 1245 strawberries in the tragedy, he has 5421 of them now. Note that all the leading zeros are omitted. That means if the number ends with a zero, the zero is lost by reversing (e.g. 1200 gives 21). Also note that the reversed number never has any trailing zeros. 
ACM needs to calculate with reversed numbers. Your task is to add two reversed numbers and output their reversed sum. Of course, the result is not unique because any particular number is a reversed form of several numbers (e.g. 21 could be 12, 120 or 1200 before reversing). Thus we must assume that no zeros were lost by reversing (e.g. assume that the original number was 12).
Input
The input consists of N cases. The first line of the input contains only positive integer N. Then follow the cases. Each case consists of exactly one line with two positive integers separated by space. These are the reversed numbers you are to add.
Output
For each case, print exactly one line containing only one integer - the reversed sum of two reversed numbers. Omit any leading zeros in the output.
Sample Input

3
24 1
4358 754
305 794

Sample Output

34
1998
1

Source
Central Europe 1998
문제에서 n조 데이터를 제시할 것이다. 이 수의 개위가 가장 높은 위치까지 거꾸로 제시된다. 이 수를 바로잡은 다음에 덧셈 연산을 한 다음에 역방향 수의 방법으로 덧셈을 출력해야 한다.
처음에 그 문제는 숫자를 문자열로 바꾸고 문자열로 숫자를 바꾸는 방법으로 해답을 구하는 것이 생각났지만 나중에 이렇게 생각하면 이전에 했던 1350의 문제와 같은 방법을 사용하고 다른 방법을 익히고 싶어서 고정밀 덧셈과 같은 방법으로 이 문제의 역방향 숫자 연산을 하는 것이 생각났다.쓰는 과정은 비교적 순조로운 편이다.
참고 사항:
1) 결과를 출력할 때 화의 역서수의 전도 0을 제거해야 한다.
2) 모든 용례가 처리되기 전에 초기화해야 한다.
코드(1AC):

#include <cstdio>
#include <cstdlib>
#include <cstring>

char num1[10], num2[10], result[10];

int main(void){
    int casenum, ii;
    int i, j;
    int len1, len2;
    int tmp,jinwei;

    scanf("%d", &casenum);
    for (ii = 0; ii < casenum; ii++){
        memset(result, 0, sizeof(result));
        memset(num1, 0, sizeof(num1));
        memset(num2, 0, sizeof(num2));
        scanf("%s%s", num1, num2);
        len1 = strlen(num1);
        len2 = strlen(num2);
        for (i = jinwei = 0; i < len1 && i < len2; i++){
            tmp = (num1[i] - '0') + (num2[i] - '0') + jinwei;
            result[i] = '0' + tmp % 10;
            jinwei = tmp / 10;
        }
        for (; i < len1; i++){
            result[i] += '0' + (num1[i] - '0' + jinwei) % 10;
            jinwei = (num1[i] - '0' + jinwei) / 10;
        }
        for (; i < len2; i++){
            result[i] += '0' + (num2[i] - '0' + jinwei) % 10;
            jinwei = (num2[i] - '0' + jinwei) / 10;
        }
        if (jinwei){
            result[i++] = '0' + jinwei;
        }
        result[i] = '\0';
        int flag = 0;
        int len3 = strlen(result);
        for (i = 0; i < len3; i++){
            if (result[i] == '0' && flag == 0){
                continue;
            }
            else{
                if (!flag){
                    flag = 1;
                }
                printf("%c", result[i]);
            }
        }
        printf("
");
    }
    return 0;
}