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  • 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
     
    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     :2013-9-111011.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
    ",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("
    ");
    102     }
    103     return 0;
    104 }
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  • 原文地址:https://www.cnblogs.com/kuangbin/p/3315073.html
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