hdu 1049

題目：

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.

 題意： 注意：每次休息時間就要增加1，這個不能忘記 每分鐘怕u，這樣一加有可能超出總長，一定中間加上判斷

#include<stdio.h>

int main()

{

int n,u,d,sum;

while(scanf("%d%d%d",&n,&u,&d),n+u+d)

{

int i;sum=0;

for(i=0;i<=n;)

{

i+=u;

           sum++;

            if(i>=n) break;

  i=i-d,sum++;//休息的時間

}

printf("%d\n",sum);

}

return 0;

}

解題者：楊聰穎

解題時間：2014-9-27

題目描述：有個蟲子，從n深處向上爬，每次爬u就要休息一分鐘，休息這一分鐘會滑下d，求最後爬上來需要多長時間，

解題思路：利用for循環進行循環相加，累計計算

難點：容易忽略休息一分鐘時，每次要加上那一分鐘

關鍵點：for循環的利用

體會：題目不難，但馬虎時容易出錯

備注：中間一次的if判斷不能忘記，否則會出現重複，回出現多加

源代碼

#include<stdio.h>

int main()

{

int n,u,d,sum;

while(scanf("%d%d%d",&n,&u,&d),n+u+d)

{

int i;sum=0;

for(i=0;i<=n;)

{

i+=u;

           sum++;

            if(i>=n) break;

  i=i-d,sum++;//休息的時間

}

printf("%d\n",sum);

}

return 0;

}

代碼詳細注釋