HDU4726-Kia's Calculation

Kia's Calculation

                                                                             Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                                                        Total Submission(s): 3508    Accepted Submission(s): 743

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

 

题意： 有2个合法的整数。 长度为 10^6。 数字的每一位都能移动， 但移动后的整数一定要是合法的， 即无前导零。 使得 A + B 最大

解题思路： 因为固定搭配不变

例如 要得到5 ，   （0， 5）（1，4）（2，3） （6， 9） （7，8） 互不干扰。

如果就统计 A中0~9分别出现多少个，B中0~9分别出现多少个。除了第一位外， 其他位都是要大就大。第一位不能出现0即可

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <functional>
#include <climits>

using namespace std;

#define LL long long
const int INF=0x3f3f3f3f;

int a[20],b[20],t,cas;
char s1[1000050],s2[1000050],ans[1000050];

int main()
{
    cas=1;
    scanf("%d",&t);
    while(t--)
    {
        memset(a,0,sizeof a);
        memset(b,0,sizeof b);
        scanf("%s%s",s1,s2);
        int len=strlen(s1);
        if(len==1)
        {
            printf("Case #%d: %d\n",cas++,(s1[0]-'0'+s2[0]-'0')%10);
            continue;
        }
        for(int i=0;s1[i];i++) a[s1[i]-'0']++;
        for(int i=0;s2[i];i++) b[s2[i]-'0']++;
        for(int i=0;i<len;i++)
        {
            int flag=0;
            for(int j=9;j>=0&&!flag;j--)
            {
                for(int k=0;k<=9;k++)
                {
                    if(j-k>=0)
                    {
                        if(i==0&&(k==0||j-k==0)) continue;
                        else
                        {
                            if(a[k]>0&&b[j-k]>0)
                            {
                                ans[i]=j+'0';
                                a[k]--;
                                b[j-k]--;
                                flag=1;
                                break;
                            }
                        }
                    }
                    else
                    {
                        if(a[k]>0&&b[j+10-k]>0)
                        {
                            ans[i]=j+'0';
                            a[k]--;
                            b[j+10-k]--;
                            flag=1;
                            break;
                        }
                    }
                }
            }
        }
        ans[len]='\0';
        printf("Case #%d: ",cas++);
        int i=0;
        while(ans[i]=='0'&&i<len-1) i++;
        while(i<len)
            printf("%c",ans[i++]);
        printf("\n");
    }
}