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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 xyw and n. Their meanings are shown above。



1x,y,w,n100.

 

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


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\n",t);
    }
    return 0;
}


posted on 2017-07-26 11:41  mthoutai  阅读(163)  评论(0编辑  收藏  举报