HDOJ 5417 Victor and Machine（模拟）

Victor and Machine

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 577    Accepted Submission(s): 333

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

   
   
   
   

    
    
    
    10
2664
939
   
   
   
   

 

Source

BestCoder Round #52 (div.2)


大意：一个机器，每运行x秒就要停止y秒，每运行w秒会吐出一个球，求第n个吐出来时使用了多少时间。

思路：各种少看条件，就把题想麻烦了，注意：
1.每次开启都会吐出一个球，这个刚开始没看到，发现样例都没法算。
2.运行的时间不能累加，就是x<w时，x不能算进运行时间，例如：x=2，w=3，那么运行的时候是不可能吐球的，
只能靠开启机器来吐球。那么这样就很简单了，刚开始我想复杂了，以为可以累加运行时间，哎...

ac代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#define MAXN 2555
#define INF 0xfffffff
#define MIN(a,b) a>b?b:a
using namespace std;
int main()
{
	int x,y,w,n;
	while(scanf("%d%d%d%d",&x,&y,&w,&n)!=EOF)
	{
		int num=0;
		int sum=0; 
		int ans=0;
		if(x<w)
		{
			printf("%d\n",(x+y)*(n-1));
			continue;
		}
		int s=x/w;
		while(1) 
		{
			num++;
			if(num==n)
			break;
			if(num+s>=n)
			ans=ans+w*(n-num),num=n;
			else
			ans=ans+x+y,num+=s;
			if(num==n)
			break;
		}
		printf("%d\n",ans);
	}
	return 0;
}