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.

A,B must be positive.

 

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.

样例输入

21 2112233445566778899 998877665544332211

样例输出

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

#include 
       
        #include 
        
         int main(){    int num = 1, n;    scanf("%d", &n);    while(n--)    {        char s1[1024], s2[1024];        int x[1024]={
     0}, y[1024]={
     0}, sum[1024]={
     0};        scanf("%s %s", s1, s2);        int len1 = strlen(s1);        int len2 = strlen(s2);        int i, k=0, j=0, p=0;        for(i=len1-1; i>=0; i--)            x[k++] = s1[i] - '0';        for(i=len2-1; i>=0; i--)            y[j++] = s2[i] - '0';        int max = k>j?k:j;        for(i=0; i<=max; i++)        {            sum[i] = x[i] + y[i] + p;            if(sum[i] > 9)            {                sum[i] -= 10;                p = 1;            }            else p=0;        }        printf("Case %d:\n", num++);        printf("%s + %s = ", s1, s2);        if(sum[max]) printf("%d", sum[max]);        for(i=max-1; i>=0; i--)            printf("%d", sum[i]);        printf("\n");    }    return 0;}