Codeforce Round #224 Div2 B

Codeforces Round #224 (Div. 2)
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Submission	Time	Verdict
5936493	02/10/2014 11:00PM	Accepted
5936360	02/10/2014 10:32PM	Time limit exceeded on test 5

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Arthur and Alexander are number busters. Today they've got a competition.

Arthur took a group of four integers a, b, w, x (0 ≤ b < w, 0 < x < w) and Alexander took integer с. Arthur and Alexander use distinct approaches to number bustings. Alexander is just a regular guy. Each second, he subtracts one from his number. In other words, he performs the assignment: c = c - 1. Arthur is a sophisticated guy. Each second Arthur performs a complex operation, described as follows: if b ≥ x, perform the assignment b = b - x, if b < x, then perform two consecutive assignments a = a - 1; b = w - (x - b).

You've got numbers a, b, w, x, c. Determine when Alexander gets ahead of Arthur if both guys start performing the operations at the same time. Assume that Alexander got ahead of Arthur if c ≤ a.

Input

The first line contains integers a, b, w, x, c (1 ≤ a ≤ 2·10⁹, 1 ≤ w ≤ 1000, 0 ≤ b < w, 0 < x < w, 1 ≤ c ≤ 2·10⁹).

Output

Print a single integer — the minimum time in seconds Alexander needs to get ahead of Arthur. You can prove that the described situation always occurs within the problem's limits.

Sample test(s)

Input

4 2 3 1 6

Output

Input

4 2 3 1 7

Output

Input

1 2 3 2 6

Output

Input

1 1 2 1 1

Output

 1 #pragma comment(linker,"/STACK:102400000,102400000")
 2 #include <cstdio>
 3 #include <vector>
 4 #include <cmath>
 5 #include <queue>
 6 #include <cstring>
 7 #include <iostream>
 8 #include <algorithm>
 9 using namespace std;
10 #define INF 0x7fffffff 
11 #define mod 1000000007
12 #define ll long long
13 #define maxn 1000025
14 #define pi acos(-1.0)
15 ll n, m, ans;
16 ll a, b, c, w, x, t, k, d, tmp;
17 int main(){
18     cin >> a >> b >> w >> x >> c;
19     m = max(0LL, c - a);
20     tmp = b;
21     while (tmp != b || d == 0){
22         k++;
23         if (b >= x){
24             b -= x;
25             d++;
26         }
27         else b = w - (x - b);
28     }
29     t = max(0LL, m / d - 1);
30     m -= t*d;
31     ans += t*k;
32     while (m > 0){
33         ans++;
34         if (b >= x){
35             b -= x;
36             m--;
37         }
38         else b = w - (x - b);
39     }
40     cout << ans;
41     return 0;
42 }

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posted @ 2014-02-11 03:47 HaibaraAi 阅读(118) 评论(0) 收藏举报

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Be careful: there is 50 points penalty for submission which fails the pretests or resubmission (except failure on the first test, denial of judgement or similar verdicts). "Passed pretests" submission verdict doesn't guarantee that the solution is absolutely correct and it will pass system tests.