BestCoder Round #52 (div.2)HDU5417 Victor and Machine-优快云博客

Victor and Machine

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 582 Accepted Submission(s): 335

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)

出题人：

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-1)/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。

#include
   
    
int main()
{
    int x,y,w,n;
    while(~scanf("%d%d%d%d",&x,&y,&w,&n))
    {
        int a=x/w+1;
        int sum=(n-1)/a*(x+y)+(n-1)%a*w;
        printf("%d\n",sum);
    }
    return 0;
}