POJ—2709—Painter—【贪心】

Painter

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3157		Accepted: 1962

Description

The local toy store sells small fingerpainting kits with between three and twelve 50ml bottles of paint, each a different color. The paints are bright and fun to work with, and have the useful property that if you mix X ml each of any three different colors, you get X ml of gray. (The paints are thick and "airy", almost like cake frosting, and when you mix them together the volume doesn't increase, the paint just gets more dense.) None of the individual colors are gray; the only way to get gray is by mixing exactly three distinct colors, but it doesn't matter which three. Your friend Emily is an elementary school teacher and every Friday she does a fingerpainting project with her class. Given the number of different colors needed, the amount of each color, and the amount of gray, your job is to calculate the number of kits needed for her class.

Input

The input consists of one or more test cases, followed by a line containing only zero that signals the end of the input. Each test case consists of a single line of five or more integers, which are separated by a space. The first integer N is the number of different colors (3 <= N <= 12). Following that are N different nonnegative integers, each at most 1,000, that specify the amount of each color needed. Last is a nonnegative integer G <= 1,000 that specifies the amount of gray needed. All quantities are in ml. 

Output

For each test case, output the smallest number of fingerpainting kits sufficient to provide the required amounts of all the colors and gray. Note that all grays are considered equal, so in order to find the minimum number of kits for a test case you may need to make grays using different combinations of three distinct colors.

Sample Input

3 40 95 21 0
7 25 60 400 250 0 60 0 500
4 90 95 75 95 10
4 90 95 75 95 11
5 0 0 0 0 0 333
0

Sample Output

2
8
2
3
4

首先除了灰色，其他颜色需要多少组颜料，然后求出辣么多组，每个颜料的余量，用剩余的去陪灰色

然后贪心策略：

每次取最多的三种颜色各1ml，配出1ml的灰色，

然后排序，重复上述过程，如果最多的三种颜色中，有一个为0，那么就要加一组，

然后所有颜色加上50ml，再重复上述过程，直到配出要求的那么多灰色

上代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
using namespace std;
int colors[20];

bool cmp(int a,int b)   //从大到小排序
{
    return a>b;
}

int main()
{
    freopen("1003.in","r",stdin);
//    freopen("1003.out","w",stdout);
    int N,G;
    while(scanf("%d",&N)!=EOF)
    {
        if(N==0)
            break;
        int countn=0;
        for(int i=0;i<N;i++)
            scanf("%d",colors+i);
        scanf("%d",&G);

        sort(colors,colors+N,cmp);
        if(colors[0]%50==0)   //排序之后，colors[0]是最多的，如果能整除50，就那么多组
            countn=colors[0]/50;
        else
            countn=colors[0]/50+1;  //如果不能整除50，要加上一组，整除直接舍弃小数位

        for(int i=0;i<N;i++)    //求出各个颜色的余量
            colors[i]=countn*50-colors[i];

        while(G>0)
        {
            sort(colors,colors+N,cmp);
            if(!colors[0] || !colors[1] || !colors[2])  //如果最多的三个颜色中，有一个用光了，但此时灰色还没配齐，就要加一组
            {
                countn++;
                for(int i=0;i<N;i++)    //加了一组，所有颜色都要加上50ml
                    colors[i]+=50;
            }

            G--;    //每次配1ml
            colors[0]--;    //最多的用掉1ml
            colors[1]--;    //次多的用掉1ml
            colors[2]--;    //第三多的用掉1ml
        }
        printf("%d\n",countn);
    }
    return 0;
}