2013南京站J题||hdu 4811 思维

http://acm.hdu.edu.cn/showproblem.php?pid=4811

Problem Description

Jenny likes balls. He has some balls and he wants to arrange them in a row on the table.
Each of those balls can be one of three possible colors: red, yellow, or blue. More precisely, Jenny has R red balls, Y yellow balls and B blue balls. He may put these balls in any order on the table, one after another. Each time Jenny places a new ball on the table, he may insert it somewhere in the middle (or at one end) of the already-placed row of balls.
Additionally, each time Jenny places a ball on the table, he scores some points (possibly zero). The number of points is calculated as follows:
1.For the first ball being placed on the table, he scores 0 point.
2.If he places the ball at one end of the row, the number of points he scores equals to the number of different colors of the already-placed balls (i.e. expect the current one) on the table.
3.If he places the ball between two balls, the number of points he scores equals to the number of different colors of the balls before the currently placed ball, plus the number of different colors of the balls after the current one.
What's the maximal total number of points that Jenny can earn by placing the balls on the table?

 

Input

There are several test cases, please process till EOF.
Each test case contains only one line with 3 integers R, Y and B, separated by single spaces. All numbers in input are non-negative and won't exceed 10 ⁹.

 

Output

For each test case, print the answer in one line.

 

Sample Input

  

   2 2 2
3 3 3
4 4 4
  

 

Sample Output

  

   15
33
51
  
  


  

解题思路：我的方法是先把x,y,z三个数的0~2的组合列出来，然后多于2的部分每增加一可以取得分数为一定的。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
ll f[3],k;
ll sum[3][3][3];
int main()
{
    sum[0][0][0]=0;
    sum[0][0][1]=0;
    sum[0][0][2]=1;
    sum[0][1][1]=1;
    sum[0][1][2]=3;
    sum[0][2][2]=6;
    sum[1][1][1]=3;
    sum[1][1][2]=6;
    sum[1][2][2]=10;
    sum[2][2][2]=15;
    while(~scanf("%I64d%I64d%I64d",&f[0],&f[1],&f[2]))
    {
        ll n=0,m=0;
        sort(f,f+3);
        for(int i=0;i<3;i++)
        {
            if(f[i]>2)
                k=2;
            else
                k=f[i];
            n+=f[i]-k,f[i]=k;
            m+=f[i];
        }
        printf("%I64d\n",sum[f[0]][f[1]][f[2]]+n*m);
    }
    return 0;
}

Problem Description

Jenny likes balls. He has some balls and he wants to arrange them in a row on the table.
Each of those balls can be one of three possible colors: red, yellow, or blue. More precisely, Jenny has R red balls, Y yellow balls and B blue balls. He may put these balls in any order on the table, one after another. Each time Jenny places a new ball on the table, he may insert it somewhere in the middle (or at one end) of the already-placed row of balls.
Additionally, each time Jenny places a ball on the table, he scores some points (possibly zero). The number of points is calculated as follows:
1.For the first ball being placed on the table, he scores 0 point.
2.If he places the ball at one end of the row, the number of points he scores equals to the number of different colors of the already-placed balls (i.e. expect the current one) on the table.
3.If he places the ball between two balls, the number of points he scores equals to the number of different colors of the balls before the currently placed ball, plus the number of different colors of the balls after the current one.
What's the maximal total number of points that Jenny can earn by placing the balls on the table?

 

Input

There are several test cases, please process till EOF.
Each test case contains only one line with 3 integers R, Y and B, separated by single spaces. All numbers in input are non-negative and won't exceed 10 ⁹.

 

Output

For each test case, print the answer in one line.

 

Sample Input

   

    2 2 2
3 3 3
4 4 4
   

 

Sample Output

   

    15
33
51