A + B Problem II

Problem Description

I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

 

 

Input

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.

 

 

Output

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.

 

 

Sample Input

2 1 2 112233445566778899 998877665544332211

 

 

Sample Output

Case 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110

#pragma warning(disable:4996)
#include<map>
#include<string>
#include<cstdio>
#include<bitset>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;

const int maxn = 1005;

char s1[maxn], s2[maxn], ans[maxn];

void sum(char *s1, char *s2) {
    int len1 = strlen(s1);
    int len2 = strlen(s2);
    for (int i = 0; i < len1; i++) s1[i] -= '0';
    for (int i = 0; i < len2; i++) s2[i] -= '0';
    int mid1 = len1 >> 1;
    int mid2 = len2 >> 1;
    /*调换位置*/
    for (int i = 0; i < mid1; i++) swap(s1[i], s1[len1 - i - 1]);
    for (int i = 0; i < mid2; i++) swap(s2[i], s2[len2 - i - 1]);
    int len3 = max(len1, len2);
    int x = 0, y;
    /*多考虑一位*/
    for (int i = 0; i <= len3; i++) {
        y = s1[i] + s2[i] + x;
        ans[i] = y % 10;
        x = y / 10;    
    }

    while (len3 > 0 && ans[len3] == 0) len3--;
    if (len3 == 0) { printf("%d\n", ans[0]); return; }
    else {
        for (int i = len3; i >= 0; i--) printf("%d", ans[i]);
        printf("\n");
        return;
    }
}

int main()
{
    int T; cin >> T;
    for (int i = 1; i <= T; i++) {
        memset(s1, 0, sizeof(s1));
        memset(s2, 0, sizeof(s2));
        scanf("%s%s", s1, s2);
        printf("Case %d:\n", i);
        printf("%s + %s = ", s1, s2);
        sum(s1, s2);
        if (i != T) printf("\n");
    }
    return 0;
}