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最新推荐文章于 2017-10-10 16:27:21 发布

转载最新推荐文章于 2017-10-10 16:27:21 发布 · 119 阅读

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原文链接：http://www.cnblogs.com/eric-blog/archive/2012/08/19/2646874.html

本文探讨了如何使用数学方法解决在玩具商店购买不同颜色手指绘画套装的问题，确保提供足够的彩色和灰色颜料给小学教师Emily的班级。通过计算不同颜色和灰度的配比，最小化所需的套装数量。

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Painter

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2422		Accepted: 1509

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

Source

Mid-Central USA 2005

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int N;
int color[15];
int rlt;

bool cmp(int a, int b)
{
    return a>b;
}

int main()
{
    int maxVolume;

    while(scanf("%d",&N)!=EOF && N)
    {
        maxVolume=0;
        for(int i=1; i<=N; i++)
        {
            scanf("%d",&color[i]);
            if(maxVolume<color[i])
            {
                maxVolume=color[i];
            }
        }
        scanf("%d",&color[0]);
        rlt=maxVolume/50;
        if(maxVolume%50!=0)
        {
            rlt++;
        }
        for(int i=1;i<=N;i++)
        {
            color[i]=rlt*50-color[i];
        }
        while(color[0]>0)
        {
            sort(color+1, color+N+1, cmp);
            if(color[3]==0)
            {
                rlt++;
                for(int i=1;i<=N;i++)
                {
                    color[i]+=50;
                }
            }
            for(int i=0;i<=3; i++)
            {
                color[i]--;
            }
        }
        printf("%d\n",rlt);
    }
    return 0;
}

转载于:https://www.cnblogs.com/eric-blog/archive/2012/08/19/2646874.html