hdu 5417 Victor and Machine 模拟_hdu5417-优快云博客

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本文探讨了关于机器定时弹出球体的问题，包括机器的启动、周期性弹出及维护周期，通过输入参数x、y、w和n计算第n个球弹出的时间。

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Victor and Machine

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 263 Accepted Submission(s): 151

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

x ,

y ,

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

Source

BestCoder Round #52 (div.2)

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

int main(){
    int x,y,w,n;
    while(scanf("%d%d%d%d",&x,&y,&w,&n)!=EOF){
        int t = 0,be = 0;
        for(t = 0;n > 0; t++){
            if(t == be){
                n--;
            }
            else if(t >= be && (t-be)%w ==0){
                n--;
            }
            if(t - be == x){
                be = t+y;
            }
        }
        printf("%d\n",t-1);

    }
    return 0;
}