HDU 1049 模拟

HDU 1049

Climbing Worm

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 16415 Accepted Submission(s): 11185



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

10 2 1
20 3 1
0 0 0

[align=left]Sample Output[/align]

17
19

[align=left]Source[/align]
East Central North America 2002

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模拟法（数据量大时TLE）

#include <iostream>

using namespace std;

int main()
{
int dis,x,y;
while(cin>>dis>>x>>y)
{
if(dis==0&&x==0&&y==0)
break;
int day=0;
while(1)
{
dis-=x;
day++;
if(dis<=0)
break;
dis+=y;
day++;
}
cout<<day<<endl;
}
}

博弈思想卡关键点

#include <iostream>

using namespace std;

int main()
{
int dis,x,y;
while(cin>>dis>>x>>y)
{
if(dis==0&&x==0&&y==0)
break;
dis-=x;
int day=2*(dis/(x-y));
dis=dis%(x-y);
dis+=x;
if(dis<=x)
cout<<day+1<<endl;
else
cout<<day+3<<endl;
}
}