Peter's Smokes

最新推荐文章于 2021-08-19 16:00:59 发布

原创最新推荐文章于 2021-08-19 16:00:59 发布 · 491 阅读

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本文介绍了如何利用烟头制作新的香烟，给出了一种贪心算法的解题方法，通过输入参数n（香烟数量）和k（制作新烟所需的烟头数量），计算Peter最多能抽到多少支烟。

Peter’s Smokes

DESCRIPTION
Peter has n cigarettes. He smokes them one by one keeping all the butts. Out of k>1 butts he can roll a new cigarette.How many cigarettes can Peter have?

Input
Input is a sequence of lines. Each line contains two integer numbers giving the values of n and k. The input is terminated by end of file.

Output
For each line of input, output one integer number on a separate line giving the maximum number of cigarettes that Peter can have.

Sample input
4 3
10 3
100 5

Sample output
5
14
124

题解：用烟头换烟，求能抽到的最多的烟数
解题方法：贪心算法

#include <cstdio>
#include <iostream>
using namespace std;

int main()
{
    int n,k,sum;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        sum=n;
        while(n/k)       //贪心算法
        {
            sum+=n/k;
            n=n/k+n%k;//n%k为不能被整除的余数用来和新增的n/k相加
        }
        printf("%d\n",sum);
    }
    return 0;
}

直接算法：

#include<cstdio>
using namespace std;
int main()
{
    int m,n;
    while(scanf("%d %d",&m,&n)!=EOF)
    {
        int sum=0;
        while(m>=n)//烟蒂达到可换烟的数目时
        {
            sum=sum+(m-m%n);//可以用烟蒂换烟的烟数
            m=(m-m%n)/n+m%n;//总烟蒂数

        }
        sum=sum+m;//烟数和不能换烟的烟蒂数
        printf("%d\n",sum);
    }
    return 0;
}