Victor and Machine
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others) Total Submission(s): 1224 Accepted Submission(s): 676
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
Source
BestCoder Round #52 (div.2)
Recommend
hujie
题目理解了，自然而然就会做了。题目大意：机器每次开启都会弹出一个球，工作状态下每过w秒弹出一个球，每工作x秒休息y秒
所以分两种情况讨论，一种是x<w，w永远不会出现在工作状态下，所以直接输出
第二种是x>=w, 求解出工作状态下可以弹出几个球+非工作状态下要弹出几个球
代码如下：
#include<cstdio>
#include<cstring>
#include<cctype>
#include<algorithm>
#include<set>
#include<cstring>
#include<string>
#include<iostream>
using namespace std;
int main (){
int x,y,w,n;
while(~scanf("%d %d %d %d",&x,&y,&w,&n)){
if(x<w)
{
printf("%d\n",(n-1)*(x+y));
continue;
}
else {
int a=(n-1)/(x/w+1)*(x+y);
int b=(n-1)%(x/w+1)*w;
printf("%d\n",a+b);
}
}
}
