A + B Problem II（1002）

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

21 2112233445566778899 998877665544332211

Sample Output

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

 哈哈哈，简不简单？就是计算a+b，别傻了(= =)，看看第二个实例就知道了，所以要用字符串模拟加法，不过呢，在这里，我还是要提一下python，2333，10^1000，完全不在话下～ 两行代码搞定2333，不过还是用C做吧2333，我怎么那么想笑呢2333

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int N = 1e3 + 5;char a[N],b[N],c[N];void add(){    int d[N];    int p = strlen(a)-1,q = strlen(b)-1;    int i = 0,left = 0;    while(p >= 0 || q >= 0)    {        d[i] = left;        if(p >= 0)            d[i] += a[p--] - '0';        if(q >= 0)            d[i] += b[q--] - '0';        left = d[i] / 10;        d[i] %= 10;        i ++;    }    if(left > 0)        d[i++] = left;    int k = 0;    while(i > 0)        c[k++] = d[--i] + '0';    c[k] = '\0';}int main(){    int t;    cin >> t;    for(int i = 1; i <= t; i++)    {        scanf("%s%s",a,b);        printf("Case %d:\n",i);        add();        printf("%s + %s = %s\n",a,b,c);        if(i < t)            printf("\n");    }    return 0;}