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
", t); continue;
}
cnt++; //忘写，WA一次
while (cnt < n) {
t += w;
cnt++;
}
printf ("%d
", t);
}
return 0;
}