Kia's Calculation hdu4726

Kia's Calculation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1367    Accepted Submission(s): 327

Problem Description

Doctor Ghee is teaching Kia how to calculate the sum of two integers. But Kia is so careless and alway forget to carry a number when the sum of two digits exceeds 9. For example, when she calculates 4567+5789, she will get 9246, and for 1234+9876, she will get 0. Ghee is angry about this, and makes a hard problem for her to solve:
Now Kia has two integers A and B, she can shuffle the digits in each number as she like, but leading zeros are not allowed. That is to say, for A = 11024, she can rearrange the number as 10124, or 41102, or many other, but 02411 is not allowed.
After she shuffles A and B, she will add them together, in her own way. And what will be the maximum possible sum of A "+" B ?

Input

The rst line has a number T (T <= 25) , indicating the number of test cases.
For each test case there are two lines. First line has the number A, and the second line has the number B.
Both A and B will have same number of digits, which is no larger than 10⁶, and without leading zeros.

Output

For test case X, output "Case #X: " first, then output the maximum possible sum without leading zeros.

Sample Input

1

5958

3036

Sample Output

Case #1: 8984

 #include <iostream>
 #include <string.h>
 #include <stdio.h>
 #include <math.h>
 #include <algorithm>
 using namespace std;
 int a[],b[];
 int fun()
 {
     int i,j,k;
     for(k=; k>=; k--)
     {
         for(i=; i>=; i--)
         {
             for(j=; j>=; j--)
             {
                 if(a[i]&&b[j]&&(i+j)%==k)
                 {
                     a[i]--,b[j]--;
                     printf("%d",k);
                     return k;
                 }
             }
         }
     }
 }
 int main()
 {
     int t,i,j,k,r;
     char x;
     scanf("%d",&t);
     getchar();
     for(r=; r<=t; r++)
     {
         memset(a,,sizeof(a));
         memset(b,,sizeof(b));
         while(x=getchar())
         {
             if(x=='\n')break;
             a[x-'']++;
         }
         while(x=getchar())
         {
             if(x=='\n')break;
             b[x-'']++;
         }
         printf("Case #%d: ",r);
         if(fun())
             for(k=; k>=; k--)
             {
                 for(i=; i>=; i--)
                 {
                     for(j=; j>=; j--)
                     {
                         while(a[i]&&b[j]&&(i+j)%==k)
                         {
                             a[i]--,b[j]--;
                             printf("%d",k);
                         }
                     }
                 }
             }
         puts("");
     }
 }