Kia’s Calculation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 83    Accepted Submission(s): 16

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  

Recommend zhuyuanchen520  

想了很久，

最后其实就是贪心构造。

最高位特殊处理。

然后后面就是不断尽量构造和大的

  1 /* ***********************************************
  2 Author        :kuangbin
  3 Created Time  :2013-9-11 12:30:33
  4 File Name     :2013-9-11\1011.cpp
  5 ************************************************ */
  6 
  7 #include <stdio.h>
  8 #include <string.h>
  9 #include <iostream>
 10 #include <algorithm>
 11 #include <vector>
 12 #include <queue>
 13 #include <set>
 14 #include <map>
 15 #include <string>
 16 #include <math.h>
 17 #include <stdlib.h>
 18 #include <time.h>
 19 using namespace std;
 20 
 21 int a[20];
 22 int b[20];
 23 
 24 char A[2000020],B[2000020];
 25 int num1[2000020],num2[2000020];
 26 int ans[2000020];
 27 
 28 
 29 int main()
 30 {
 31     //freopen("in.txt","r",stdin);
 32     //freopen("out.txt","w",stdout);
 33     int T;
 34     int iCase = 0;
 35     scanf("%d",&T);
 36     while(T--)
 37     {
 38         iCase++;
 39         scanf("%s%s",A,B);
 40         int n = strlen(A);
 41         for(int i = 0;i < n;i++)
 42         {
 43             num1[i] = A[i] - '0';
 44             num2[i] = B[i] - '0';
 45         }
 46         if(n == 1)
 47         {
 48             printf("Case #%d: %d\n",iCase,(num1[0]+num2[0])%10);
 49             continue;
 50         }
 51         memset(a,0,sizeof(a));
 52         memset(b,0,sizeof(b));
 53         for(int i = 0;i < n;i++)
 54         {
 55             a[num1[i]] ++;
 56             b[num2[i]] ++;
 57         }
 58         int x = 0, y = 0;
 59         int ttt = -1;
 60         for(int i = 1;i <= 9;i++)
 61             for(int j = 1;j <= 9;j++)
 62                 if(a[i] && b[j] && ((i+j)%10) > ttt )
 63                 {
 64                     x = i;
 65                     y = j;
 66                     ttt = (x+y)%10;
 67                 }
 68         a[x]--;
 69         b[y]--;
 70         int cnt = 0;
 71         ans[cnt++] = (x+y)%10;
 72 
 73         for(int p = 9;p >= 0;p--)
 74         {
 75             for(int i = 0;i <= 9;i++)
 76                 if(a[i])
 77                 {
 78                     if(i <= p)
 79                     {
 80                         int j = p-i;
 81                         int k = min(a[i],b[j]);
 82                         a[i] -= k;
 83                         b[j] -= k;
 84                         while(k--)
 85                             ans[cnt++] = p;
 86                     }
 87                     int j = 10 + p - i;
 88                     if(j > 9)continue;
 89                     int k = min(a[i],b[j]);
 90                     a[i] -= k;
 91                     b[j] -= k;
 92                     while(k--)
 93                         ans[cnt++] = p;
 94                 }
 95         }
 96         printf("Case #%d: ",iCase);
 97         int s = 0;
 98         while(s < cnt-1 && ans[s] == 0)s++;
 99         for(int i = s;i < cnt;i++)
100             printf("%d",ans[i]);
101         printf("\n");
102     }
103     return 0;
104 }