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：
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 }