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 10 ⁶, 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,'\0',sizeof a);

26    memset(b,'\0',sizeof b);

27    memset(c,'\0',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("%c\n",c[0]);

76   }

77   return 0;

78 }

View Code