HDU 1049Climbing Worm 解题报告

Description

An inch worm is at the bottom of a well n inches deep. It has enough energy to climb u inches every minute, but then has to rest a minute before climbing again. During the rest, it slips down d inches. The process of climbing and resting then repeats. How long before the worm climbs out of the well? We'll always count a portion of a minute as a whole minute and if the worm just reaches the top of the well at the end of its climbing, we'll assume the worm makes it out. 

Input

There will be multiple problem instances. Each line will contain 3 positive integers n, u and d. These give the values mentioned in the paragraph above. Furthermore, you may assume d < u and n < 100. A value of n = 0 indicates end of output. 

Output

Each input instance should generate a single integer on a line, indicating the number of minutes it takes for the worm to climb out of the well. 

Sample Input

10 2 120 3 10 0 0

Sample Output

1719

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题目中说蠕虫身长1英寸，这条信息并没什么用。

小学脑筋急转弯题，利用计算机计算的便利很容易就能实现一个O（n）的算法。

另外整理的时候想到一个O（1）的算法，hhhh。

先说O（n）的算法。

#include<stdio.h>int main(){    int n,u,d;    while(scanf("%d%d%d",&n,&u,&d)==3&&n)    {        int sum=0,minute=0;        while(sum<n)        {            sum+=u;            minute++;            if(sum>=n)            {                                break;            }else{                sum-=d;                minute++;            }        }        printf("%d\n",minute);    }    return 0;}

整理的时候想到的O（1）的算法。

#include<stdio.h>int main(){    int n,u,d;    while(scanf("%d%d%d",&n,&u,&d)==3&&n)    {        int sum,minute,num;        num=n/(u-d);        sum=(u-d)*num;        minute=num*2;        int flag=n-u;        num=(sum-flag)/(u-d);        sum-=num*(u-d);        minute-=2*num;        if(n-sum<=d)            minute--;        else            minute++;        printf("%d\n",minute);    }    return 0;}