BestCoder Round #52 (div.2) 1001题

Victor and Machine

Accepts: 452

Submissions: 1123

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 131072/65536 K (Java/Others)

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。

$1\le x,y,w,n\le 100$.

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 3 398 76 54 3210 9 8 100

Sample Output

102664939

分析：理解题意的时候总是想不明白，想了好久才明白，因为每次重新开始都会弹出一个球，而且每次修理后都会重新开始，理解这些就好了，然后直接模拟就好，1A

代码：

#include <cstdio>#include <cstring>#include <algorithm>#include<iostream>using namespace std;#define LL long longint main(){    int x,y,w,n,a,b;    while(cin>>x>>y>>w>>n){         a=x/w;　／／在要停下修理之前可以弹出的小球个数        b=x%w;        int sum=0,t=0;        n--;        int i;       for( i=n;i>0;i--){　／／每次开始－－，表示弹一个球            if(i-a<=0){                sum+=i*w;／／有可能不用弹ａ个就达到ｎ个要求了            break;            }            else{            i-=a;            sum+=a*w+y+b;／／时间和加上从开始一直到再一次的开始            }       }        cout<<sum<<endl;    }    return 0;}