HDU 5417 Victor and Machine(模拟||递推)

Victor and Machine

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 253 Accepted Submission(s): 144

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
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    //第一次参加bestcoder 呵呵 太渣了我
   
   
   
   
   
   
   
   

    
    
    
    //递推
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<stdlib.h>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#define exp 1e-10
using namespace std;
const int N = 16;
const int inf = 2147483647;
const int mod = 2009;
int main()
{
    int x,y,w,n,s,t;
    while(~scanf("%d%d%d%d",&x,&y,&w,&n))
    {
        s=x/w+1;
        if(s==1)
            printf("%d\n",(n-1)*(x+y));
        else if(n%s==0)
        {
            t=n/s-1;
            printf("%d\n",t*(x+y)+(s-1)*w);
        }
        else
        {
            t=n/s;
            printf("%d\n",t*(x+y)+(n-t*s-1)*w);
        }
    }
    return 0;
}