A + B Problem II---hdu1002

 

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 260647    Accepted Submission(s): 50397

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

 

 

 

#include<stdio.h>
#include<string.h>
#define as 1000
int main()
{
    int as1[as+20],as2[as+20],cot=0,j,t;
    char shuru1[as+20],shuru2[as+20];
    int n,i;
    scanf("%d",&n);
    getchar();//吸收回车符
    t=n;
    while(n--)
    {
        cot++;
        scanf("%s",shuru1);
        scanf("%s",shuru2);
        memset(as1,0,sizeof(as1));
        memset(as2,0,sizeof(as2));
        for(i=0,j=strlen(shuru1)-1;j>=0;j--,i++)
        {
            as1[i]=shuru1[j]-'0';
            
        }
        for(i=0,j=strlen(shuru2)-1;j>=0;j--,i++)
        {
            as2[i]=shuru2[j]-'0';
        }
        for(i=0;i<as+20;i++)
        {
            as1[i]+=as2[i];
            if(as1[i]>=10)//判断是否满十进一
            {
                as1[i+1]++;
                as1[i]-=10;
            }
            
        }
        for(i=as+19;(i>=0)&&(as1[i]==0);i--);//去掉结果前面多余的0
        printf("Case %d:\n",cot);
        printf("%s + %s = ",shuru1,shuru2);
         if(i>=0)
         {
             for(;i>=0;i--)
             printf("%d",as1[i]);
         }
         else
         printf("0");
         printf("\n");
         if(cot!=t)
         printf("\n");//最后不要多换行
    }
    return 0;
}