HDUOJ---------Kia's Calculation

Kia's Calculation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 624    Accepted Submission(s): 178

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

 

Source

2013 ACM/ICPC Asia Regional Online —— Warmup2

代码：786ms 采用数组存储，来做的...比较惊险的，卡过去了！！

代码：

 #include<iostream>
 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<algorithm>
 #define maxn 1000000  //1e6
 using namespace std;
 char a[maxn+5],b[maxn+5],c[maxn+5];
 void cal(int *str,char *a,int len)
 {
    for(int i=0;i<len;i++)
    {
      str[a[i]-'0']++;
    }
 }
 int main()
 {
   int t,i,j,count=1,max,posi,posj,len1,k;
   //freopen("test.in","r",stdin);
   //freopen("test.out","w",stdout);
   scanf("%d",&t);
   while(t--)
   {
    int A[10]={0},B[10]={0},flag=0;
    memset(a,'\0',sizeof a);
    memset(b,'\0',sizeof b);
    memset(c,'\0',sizeof c);
    scanf("%s%s",a,b);
    len1=strlen(a);
    cal(A,a,len1);
    cal(B,b,len1);
    printf("Case #%d: ",count++);
    for(k=1;k<=len1;k++)
    {
       max=-1;
      for(i=9;i>=0;i--)
      {
          if(k==1&&i==0&&len1!=1)  continue;
          if(A[i])
          {
          for(j=9;j>=0;j--)
          {
            if(k==1&&j==0&&len1!=1)  continue;
             if(B[j])
             {
                int temp=(i+j)%10;
                if(max<temp)
                {
                   max=temp;
                   posi=i;
                   posj=j;
                }
             }
           if(max==9)break;
          }
         }
           if(max==9)break;
      }
      A[posi]--;
      B[posj]--;
      c[flag++]=max+'0';
    }
    if(flag>1)
    {
     if(c[0]>'0')
        printf("%c",c[0]);
     for(i=1;i<flag;i++)
     {
       printf("%c",c[i]);
       if(c[0]<='0'&&c[i]=='0') break;
     }
      puts("");
   }
   else
       printf("%c\n",c[0]);
   }
   return 0;
 }