CF349B Color the fence-20140208

最新推荐文章于 2021-09-07 19:41:04 发布

原创最新推荐文章于 2021-09-07 19:41:04 发布 · 869 阅读

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水题堆同时被 2 个专栏收录

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贪心

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本文通过实例解析贪心算法在解决涂栅栏问题中的应用，详细讲解了算法思路、实现步骤以及代码实现，旨在帮助读者理解并掌握此类问题的解决方法。

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题目描述

Color the fence

时间限制：1000 ms | 内存限制：65535 KB

难度：2

描述

Tom has fallen in love with Mary. Now Tom wants to show his love and write a number on the fence opposite to

Mary’s house. Tom thinks that the larger the numbers is, the more chance to win Mary’s heart he has.

Unfortunately, Tom could only get V liters paint. He did the math and concluded that digit i requires ai liters paint.

Besides,Tom heard that Mary doesn’t like zero.That’s why Tom won’t use them in his number.

Help Tom find the maximum number he can write on the fence.

输入

There are multiple test cases.
Each case the first line contains a nonnegative integer V(0≤V≤10^6).
The second line contains nine positive integers a1,a2,……，a9(1≤ai≤10^5).

输出

Printf the maximum number Tom can write on the fence. If he has too little paint for any digit, print -1.

样例输入

55 4 3 2 1 2 3 4 529 11 1 12 5 8 9 10 6

解题报告

典型的贪心算法即可解决问题。

先取最大长度，再分别取每位上的最大值。

代码

#include <iostream>
#include <stdio.h>

/*贪心算法*/

using namespace std;
int main()
{
    long a[9],minnmum=100005,v,rest,times,mtimes;
        while(~scanf("%d\n",&v))
    {
        minnmum=100005;
        for(int i=0;i<9;i++)
        {
            scanf("%d",&a[i]);
            if(a[i]<=minnmum)
                {
                minnmum=a[i];
                mtimes=i;
                }//取得最小值和最小下标
        }
        rest=v%minnmum;//取得最大字长下拉伸区间
        times=v/minnmum;//取得最大字长
        if(v<minnmum)
        {
            printf("-1\n");
            continue;
        }
        while(times!=0)
        {
           for(int o=8;o>=mtimes;o--)
               {int t = (v-a[o])/minnmum;
                if(t>=times-1&&v>=a[o])
                {
                    printf("%d",o+1);
                    times--;
                    v-=a[o];
                    break;
                }
            //找最大值*/
               }
        }
        printf("\n");
    }
}