Time Limit: 2000/1000 MS ( Java/Others) ? ?Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 481 ? ?Accepted Submission(s): 133

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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 ?

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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.

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Output

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

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Sample Input

1

5958

3036

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Sample Output

Case #1: 8984

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Source

2013 ACM/ICPC Asia Regional Online —— Warmup2

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Recommend

zhuyuanchen520

错了好多次了,用贪心的方法,从大数字到小数字来找,这样才不会超 时!

#include   
#include   
#include   
using namespace std;  
int p[2][10];  
char str1[1000050],str2[1000050];  
int main ()  
{  
    int tcase,a,b,num1,num2,i,j,k,tt=1;  
    scanf("%d",&tcase);  
    gets(str1);  
    while(tcase--)  
    {  
        gets(str1);  
        gets(str2);  
        num1=0;  
        memset(p,0,sizeof(p));  
        for(num1=0;str1[num1]!='\0';num1++)  
        {  
            p[0][str1[num1]-'0']++;  
            p[1][str2[num1]-'0']++;  
        }  
         printf("Case #%d: ",tt++);  
        bool flag=true,flag2=true;  
        int li;  
        for(i=0,li=9;i=0&&flag;li--)  
            {  
               for(j=0;j<=9&&flag;j++)  
               {  
                  k=(li-j+10)%10;  
                  if(!i&&(!j||!k))  
                  {  
                    continue;  
                  }  
                  if(p[0][j]&&p[1][k])  
                  {  
  
                      p[0][j]--;p[1][k]--;  
                      flag=false;  
                      if(flag2&&li!=0)  
                      printf("%d",li),flag2=false;  
                      else if(!flag2)  
                      printf("%d",li);  
                      goto my;  
                  }  
               }  
            }  
            my:;  
            if(!i)  
            li=9;  
        }  
        if(flag2)  
        {  
            printf("0\n");  
        }  
        else  
        printf("\n");  
  
    }  
    return 0;  
}  

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