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[10],b[10];
int fun()
{
    int i,j,k;
    for(k=9; k>=0; k--)
    {
        for(i=9; i>=1; i--)
        {
            for(j=9; j>=1; j--)
            {
                if(a[i]&&b[j]&&(i+j)%10==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=1; r<=t; r++)
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        while(x=getchar())
        {
            if(x=='
')break;
            a[x-'0']++;
        }
        while(x=getchar())
        {
            if(x=='
')break;
            b[x-'0']++;
        }
        printf("Case #%d: ",r);
        if(fun())
            for(k=9; k>=0; k--)
            {
                for(i=9; i>=0; i--)
                {
                    for(j=9; j>=0; j--)
                    {
                        while(a[i]&&b[j]&&(i+j)%10==k)
                        {
                            a[i]--,b[j]--;
                            printf("%d",k);
                        }
                    }
                }
            }
        puts("");
    }
}