Victor and Machine

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 87 Accepted Submission(s): 48

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

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)

题意：有中文版的。就不多说了

分析：机器执行关闭为一回合。计算出每回合能弹出多少个小球。然后处理一下不是执行/关闭的情况就OK了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
#define ll long long
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;

int main ()
{
    int x,y,w,n;
    while (scanf ("%d%d%d%d",&x,&y,&w,&n)==4)
    {
        int t;
        if (x < w)
            t = (x+y)*(n-1);
        else
        {
            int a=x/w+1;
            if (n%a==0)
                t = (n/a)*(x+y)-y-x%w;
            else
                t = (n/a)*(x+y)+(n%a-1)*w;
        }
        printf ("%d
",t);
    }
    return 0;
}