Victor and Machine（推公式或者模拟）

Link：http://acm.hdu.edu.cn/showproblem.php?pid=5417

Victor and Machine

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)

Total Submission(s): 84 Accepted Submission(s): 46



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)



官方题解：

1000 Victor and Machine

首先是关于题意的问题，之前有人向我提出题意不清，对于y<w的情况，没有说明该如何处理，但是在咨询了其他几个小伙伴之后，我决定还是不修改题面了，因为题中已经说明了是“机器每次开启的瞬间”，也就是说机器关闭之后w就会清零。我不大清楚是否有人因为这个坑点而被hack或者fst了，如果有的话，对此我只能表示遗憾。

因为数据范围很小，所以我们可以手动模拟，枚举每个小球弹出的时刻，再特判一下机器的关闭情况就可以了。实际处理起来的话或许有些蛋疼，但是，只要细心一点并且耐心一点，都是能够AC的。

另外一种解法就是推公式，我们发现，在机器每次开启的时间，都有x/w+1个小球被弹出（无论w是否整除x），假设这个数目为a，那么每弹出a个小球就需要花费x+y的时间，那么，第n个小球弹出的时刻就是n/a*(x+y)+(n-1)%a*w。要注意的是，这里需要用(n-1)%a*w，因为题中求的是第n个小球弹出的时刻等于是弹出n-1个小球所花费的时间，然而，只要测试一下样例就能注意到这一点了。

AC code：

#include <stdio.h>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#include<vector>
#define LL long long
#define MAXN 1000100
using namespace std;
const int mod=1000000007;
int ans,cnt,tt;
int main()
{
   	int x,y,w,n,m,t;
   	while(scanf("%d%d%d%d",&x,&y,&w,&n)!=EOF)
   	{
   		m=x/w+1;
   		n--;
		ans=(n/m)*(x+y)+(n%m)*w;	
   		printf("%d\n",ans);	
	}
    return 0;
}