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：

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<cstring>
 5 #include<algorithm>
 6 #define maxn 1000000  //1e6
 7 using namespace std;
 8 char a[maxn+5],b[maxn+5],c[maxn+5];
 9 void cal(int *str,char *a,int len)
10 {
11    for(int i=0;i<len;i++)
12    {
13      str[a[i]-'0']++;
14    }
15 }
16 int main()
17 {
18   int t,i,j,count=1,max,posi,posj,len1,k;
19   //freopen("test.in","r",stdin);
20   //freopen("test.out","w",stdout);
21   scanf("%d",&t);
22   while(t--)
23   {
24    int A[10]={0},B[10]={0},flag=0;
25    memset(a,'',sizeof a);
26    memset(b,'',sizeof b);
27    memset(c,'',sizeof c);
28    scanf("%s%s",a,b);
29    len1=strlen(a);
30    cal(A,a,len1);
31    cal(B,b,len1);
32    printf("Case #%d: ",count++);
33    for(k=1;k<=len1;k++)
34    {
35       max=-1;
36      for(i=9;i>=0;i--)
37      { 
38          if(k==1&&i==0&&len1!=1)  continue;
39          if(A[i])
40          {
41          for(j=9;j>=0;j--)
42          {
43            if(k==1&&j==0&&len1!=1)  continue;
44             if(B[j])
45             {
46                int temp=(i+j)%10;
47                if(max<temp)
48                {
49                   max=temp;
50                   posi=i;
51                   posj=j;
52                }
53             }
54           if(max==9)break;
55          }
56         }
57           if(max==9)break;
58      }
59      A[posi]--;
60      B[posj]--;
61      c[flag++]=max+'0';
62    }
63    if(flag>1)
64    {
65     if(c[0]>'0')
66        printf("%c",c[0]);
67     for(i=1;i<flag;i++)
68     {
69       printf("%c",c[i]);
70       if(c[0]<='0'&&c[i]=='0') break;
71     }
72      puts("");
73   }
74   else
75       printf("%cn",c[0]);
76   }
77   return 0;
78 }

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

Kia's Calculation