构造 BestCoder Round #52 (div.2) 1001 Victor and Machine

题目传送门

题意：有中文版的

分析：首先要知道机器关闭后，w是清零的。所以一次(x + y)的循环弹出的小球个数是固定的，为x / w + 1，那么在边界时讨论一下就行了

收获：这种题目不难，理解清楚题意，yy出可行的解法总能做出来

代码：

/************************************************
* Author        :Running_Time
* Created Time  :2015-8-22 18:55:05
* File Name     :A.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;

int main(void)    {
    int x, y, w, n;
    while (scanf ("%d%d%d%d", &x, &y, &w, &n) == 4)	{
        int cnt = 0;
        int t = 0;
        int a = x / w + 1;
        while (cnt + a <= n)	{
            cnt += a;
            if (cnt == n)	{
                t += (a - 1) * w;	break;
            }
            else if (cnt == n - 1)	{
                t += x + y;	break;
            }
			else t += (x + y);
        }
        if (cnt == n || cnt == n - 1)	{
            printf ("%d\n", t);	continue;
        }
        cnt++;		//忘写，WA一次
        while (cnt < n)	{
            t += w;
            cnt++;
        }
        printf ("%d\n", t);
    }

    return 0;
}