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

 

分析：高精度大数加法，肯定用数组来解决，注意小坑。。。

#include 
     
      #include 
      
       #define MAX_LENTH 1001int main(){    char s1[MAX_LENTH],s2[MAX_LENTH];    int a[MAX_LENTH],b[MAX_LENTH],c[MAX_LENTH];   //a用来储存第一个数，b第二个数，c是a+b；    int n,i,j,l1,l2,z;    scanf("%d",&n);    for(i=0;i
       
        =10)            {                c[j+1]+=c[j]/10;                c[j]=c[j]%10;            }        }        printf( "Case %d:\n",i+1);        printf("%s + %s = ",s1,s2);        z=0;                                      //确定首位        for(j=MAX_LENTH-1;j>=0;j--)        {            if(z==0)            {                if(c[j]!=0){                    printf("%d",c[j]);                    z=1;                }            }            else                printf("%d",c[j]);        }        if(z==0)                            //和为0            printf("0");        if(i