首页 > ACM题库 > HDU-杭电 > HDU 4726-Kia’s Calculation-贪心-[解题报告]HOJ
2015
09-17

HDU 4726-Kia’s Calculation-贪心-[解题报告]HOJ

Kia’s Calculation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 481    Accepted Submission(s): 133



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
 


Recommend
zhuyuanchen520
错了好多次了,用贪心的方法,从大数字到小数字来找,这样才不会超 时!
#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
int p[2][10];
char str1[1000050],str2[1000050];
int main ()
{
    int tcase,a,b,num1,num2,i,j,k,tt=1;
    scanf("%d",&tcase);
    gets(str1);
    while(tcase--)
    {
        gets(str1);
        gets(str2);
        num1=0;
        memset(p,0,sizeof(p));
        for(num1=0;str1[num1]!='\0';num1++)
        {
            p[0][str1[num1]-'0']++;
            p[1][str2[num1]-'0']++;
        }
         printf("Case #%d: ",tt++);
        bool flag=true,flag2=true;
        int li;
        for(i=0,li=9;i<num1;i++)
        {

            flag=true;
            for(;li>=0&&flag;li--)
            {
               for(j=0;j<=9&&flag;j++)
               {
                  k=(li-j+10)%10;
                  if(!i&&(!j||!k))
                  {
                    continue;
                  }
                  if(p[0][j]&&p[1][k])
                  {

                      p[0][j]--;p[1][k]--;
                      flag=false;
                      if(flag2&&li!=0)
                      printf("%d",li),flag2=false;
                      else if(!flag2)
                      printf("%d",li);
                      goto my;
                  }
               }
            }
            my:;
            if(!i)
            li=9;
        }
        if(flag2)
        {
            printf("0\n");
        }
        else
        printf("\n");

    }
    return 0;
}

版权声明:本文为博主原创文章,未经博主允许不得转载。

参考:http://blog.csdn.net/mengzhengnan/article/details/11599357