HDU 4726 Kia's Calculation（贪心）

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 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

 

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

 /* ***********************************************
 Author        :kuangbin
 Created Time  :2013-9-11 12:30:33
 File Name     :2013-9-11\1011.cpp
 ************************************************ */

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <set>
 #include <map>
 #include <string>
 #include <math.h>
 #include <stdlib.h>
 #include <time.h>
 using namespace std;

 int a[];
 int b[];

 char A[],B[];
 int num1[],num2[];
 int ans[];

 int main()
 {
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     int T;
     int iCase = ;
     scanf("%d",&T);
     while(T--)
     {
         iCase++;
         scanf("%s%s",A,B);
         int n = strlen(A);
         for(int i = ;i < n;i++)
         {
             num1[i] = A[i] - '';
             num2[i] = B[i] - '';
         }
         if(n == )
         {
             printf("Case #%d: %d\n",iCase,(num1[]+num2[])%);
             continue;
         }
         memset(a,,sizeof(a));
         memset(b,,sizeof(b));
         for(int i = ;i < n;i++)
         {
             a[num1[i]] ++;
             b[num2[i]] ++;
         }
         int x = , y = ;
         int ttt = -;
         for(int i = ;i <= ;i++)
             for(int j = ;j <= ;j++)
                 if(a[i] && b[j] && ((i+j)%) > ttt )
                 {
                     x = i;
                     y = j;
                     ttt = (x+y)%;
                 }
         a[x]--;
         b[y]--;
         int cnt = ;
         ans[cnt++] = (x+y)%;

         for(int p = ;p >= ;p--)
         {
             for(int i = ;i <= ;i++)
                 if(a[i])
                 {
                     if(i <= p)
                     {
                         int j = p-i;
                         int k = min(a[i],b[j]);
                         a[i] -= k;
                         b[j] -= k;
                         while(k--)
                             ans[cnt++] = p;
                     }
                     int j =  + p - i;
                     if(j > )continue;
                     int k = min(a[i],b[j]);
                     a[i] -= k;
                     b[j] -= k;
                     while(k--)
                         ans[cnt++] = p;
                 }
         }
         printf("Case #%d: ",iCase);
         int s = ;
         while(s < cnt- && ans[s] == )s++;
         for(int i = s;i < cnt;i++)
             printf("%d",ans[i]);
         printf("\n");
     }
     return ;
 }