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

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("");

     }

 }

巴特西

Kia's Calculation hdu4726

Kia's Calculation

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