原题：

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：

模拟进制，字符串处理；

原码：

#include
    
     #include
     
      int main(){    char a1[1002],a2[1002];    int c,i,j,n,m,s1[1002],s2[1002],l1,l2,count;    count=1;    scanf("%d",&n);    m=n;    while(m--)    {        memset(s1,0,sizeof(s1));        memset(s2,0,sizeof(s2));//s1、s2全部赋值为0；        scanf("%s%s",a1,a2);        l1=strlen(a1);        l2=strlen(a2);        c=0;        for(i=l1-1; i>=0; i--)        {            s1[c++]=a1[i]-'0';        }        c=0;//将字符串转换成数字        for(i=l2-1; i>=0; i--)        {            s2[c++]=a2[i]-'0';        }//将字符串转换成数字        for(i=0; i<1002; i++)        {            s1[i]+=s2[i];//逐位相加            if(s1[i]>=10)//判断是否进位            {                s1[i]-=10;                s1[i+1]++;//进位            }        }        printf("Case %d:\n",count++);        printf("%s + %s = ",a1,a2);        for(i=1001; i>=0; i--)        {            if(s1[i])                break;        }   //跳出多余的0；        for(j=i; j>=0; j--)        {            printf("%d",s1[j]);//已经跳出多余的0，依次输出。        }        printf("\n");        if(count!=n+1)            printf("\n");    }    return 0;}