hdu5417（BC） Victor and Machine

题目链接：点这儿

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

seconds. However, the machine has some flaws, every time after

seconds of process the machine has to turn off for

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

second, the machine opens for the first time. Victor wants to know when the

-th ball will be popped out. Could you tell him?

Input

The input contains several test cases, at most

cases.

Each line has four integers

. Their meanings are shown above。

Output

For each test case, you should output a line contains a number indicates the time when the

-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;
}

hdu5417（BC） Victor and Machine

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