HDOJ 5417 Victor and Machine

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探讨一个机器按周期性规律弹射小球的问题，包括机器的启动、暂停及小球弹射的时间计算。

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Problem Description

Victor has a machine. When the machine starts up, it will pop out a ball immediately. After that, the machine will pop out a ball every

w seconds. However, the machine has some flaws, every time after

x seconds of process the machine has to turn off for

y seconds for maintenance work. At the second the machine will be shut down, it may pop out a ball. And while it's off, the machine will pop out no ball before the machine restart.

Now, at the

0 second, the machine opens for the first time. Victor wants to know when the

n-th ball will be popped out. Could you tell him?

Input

The input contains several test cases, at most

100 cases.

Each line has four integers

w and

n. Their meanings are shown above。

1≤x,y,w,n≤100.

Output

For each test case, you should output a line contains a number indicates the time when the

n-th ball will be popped out.

Sample Input

2 3 3 3
98 76 54 32
10 9 8 100

Sample Output

10
2664
939

题目大意：一个机器每隔w (/s)弹射一颗小球，机器开启的瞬间(第0时刻)会有一颗小球弹出，但是每隔 x (/s)机器会关机，修整y (/s)然后重启。

很明显是一个以 ( x+y) 为周期的题目，求出每个周期会有几颗小球弹射就可以了，注意边界小球的处理。需要注意的是如果 w > x 那么w将是没有用的条件，而且时间不进行累计。

思路：这种运动是有周期的，找到单个区间的数量关系就行了，然后在处理一下余数的问题

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#define LL int
#define inf 0x3f3f3f3f
using namespace std;

int main()
{
    LL  n,m,x,w,y,i,j,cycle;
    while(~scanf("%d%d%d%d",&x,&y,&w,&n))
    {
        cycle=x/w+1;//一个周期内小球的弹出数量
        printf("%d\n",((n-1)/cycle)*(x+y)+( (n-1)%cycle )*w );
    }//求第n个小球的弹出数量=n-1个小球弹出的总时间
    return 0;
}

或者，枚举n产生的条件：

        int X=x;
        int time=0;
        for(int i=2;i<=n;i++)
        {
            if(time+w<=x)
            {
                time+=w;
            }
            else
            {
                time=x;
                time+=y;
                x=x+y+X;
            }
        }
        cout<<time<<endl;