HDOJ 5417 Victor and Machine 水
Victor and Machine
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 240 Accepted Submission(s): 134
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 the0 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?
Now, at the
Input
The input contains several test cases, at most 100 cases.
Each line has four integersx , y , w and n .
Their meanings are shown above。
Each line has four integers
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
/* *********************************************** Author :CKboss Created Time :2015年08月22日 星期六 21时57分18秒 File Name :HDOJ5417.cpp ************************************************ */ #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <string> #include <cmath> #include <cstdlib> #include <vector> #include <queue> #include <set> #include <map> using namespace std; int x,y,w,n; int main() { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); while(cin>>x>>y>>w>>n) { int X=x; int time=0; for(int i=2;i<=n;i++) { if(time+w<=x) { time+=w; } else { time=x; time+=y; x=x+y+X; } } cout<<time<<endl; } return 0; }
posted on 2017-07-31 20:50 yjbjingcha 阅读(126) 评论(0) 编辑 收藏 举报