hdu 3953

I'll play a trick on you

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 471    Accepted Submission(s): 304

Problem Description

Please look the picture carefully. Then I'll give you two integers and your task is output the third one.
Please never doubt the picture.

 

Input

The first line is a number T(1<=T<=30), represents the number of case. The next T blocks follow each indicates a case.
Each case contains two integers A,B (1<=B<=A<=10¹⁰⁰)

 

Output

For each case, output the number of case and the third integer.(as shown in the sample output)

 

Sample Input

399 7245 2739 18

 

Sample Output

Case 1: 27Case 2: 18Case 3: 21
Hint
If you have any idea to work out the ? and explain why but couldn't get Accepted , please email me ([e-mail]), the first person will get 100RMB from me.
 

 

Author

NotOnlySuccess

 

Source

2011 Alibaba Programming Contest

 

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lcy

坑爹的题，有木有！有木有！！！

#include<stdio.h>#include<string.h>int main(){    int t;    int i,j;    char a[200],b[200];    int sum;    scanf("%d\n",&t);    for(i=1;i<=t;i++)    {        scanf("%s %s",a,b);        sum=0;        for(j=0;j<strlen(a);j++)        sum+=a[j]-48;        for(j=0;j<strlen(b);j++)        sum+=b[j]-48;        printf("Case %d: %d\n",i,sum);    }    return 0;}