Overview

This is a problem where we compare the shortest time for Takahashi to arrive at the station by optimally using his full-speed dash, against the time the last train arrives at the station, to determine whether he can catch the last train.

Analysis

Problem Setup

• Takahashi: wants to move from coordinate \(0\) to coordinate \(G\)
• Last train: moving from coordinate \(M\) to coordinate \(G\) at speed \(V\)
• Takahashi has normal speed \(1\), and can dash at speed \(D\) (usable for up to \(K\) seconds)

Key Observations

The last train’s arrival time is fixed

The last train travels a distance of \(M - G\) at speed \(V\), so its arrival time is: $\(\text{Train arrival time} = \frac{M - G}{V}\)$

Finding Takahashi’s shortest arrival time

We consider the shortest time for Takahashi to travel distance \(G\). Since dashing makes him move faster, it is optimal to use as much full-speed dash as possible.

We consider two cases:

Case 1: He can arrive using only full-speed dash (\(D \times K \geq G\))

He can cover distance \(G\) using only full-speed dash. $\(\text{Shortest time} = \frac{G}{D}\)$

Case 2: Full-speed dash alone is not enough (\(D \times K < G\))

He dashes for \(K\) seconds covering \(D \times K\), then travels the remaining \(G - D \times K\) at normal speed \(1\). $\(\text{Shortest time} = K + (G - D \times K)\)$

Concrete Example

For \(G = 100\), \(D = 3\), \(K = 10\): - Distance covered by dashing: \(3 \times 10 = 30\) - Remaining distance: \(100 - 30 = 70\) - Shortest time: \(10 + 70 = 80\) seconds

Algorithm

1. Calculate the time the last train arrives at the station: \(\text{train\_time} = \frac{M - G}{V}\)
2. Calculate Takahashi’s shortest arrival time:
   • If \(D \times K \geq G\): \(\text{takahashi\_time} = \frac{G}{D}\)
   • Otherwise: \(\text{takahashi\_time} = K + (G - D \times K)\)
3. If \(\text{takahashi\_time} \leq \text{train\_time}\), output Yes; otherwise output No

Complexity

• Time complexity: \(O(1)\) (only arithmetic operations)
• Space complexity: \(O(1)\) (only variables used)

Implementation Notes

• Beware of overflow: When computing \(D \times K\), if both \(D\) and \(K\) are at their maximum of \(10^9\), the product reaches \(10^{18}\). In Python there is no integer overflow so this is not an issue, but care is needed in other languages.

• Floating-point comparison: This problem involves comparing results of division, but since the inputs are integers, it is possible to transform the inequality and compare using integers only (if a more precise implementation is needed).

• Optimal use of full-speed dash: Since there is no reason to rest midway, the strategy of “using all full-speed dash at the beginning” is optimal.

  Source Code

G, M, D, K, V = map(int, input().split())

# 終電が駅に到着する時刻
# 終電は座標Mから駅(座標G)へ向かう。距離は M - G
train_time = (M - G) / V

# 高橋君が駅に到着する最短時間
# 全力ダッシュをK秒使うと、その間は速度D、残りは速度1
# 全力ダッシュで進める距離: D * min(K, t) （t秒かけるとき、最大K秒まで全力ダッシュ可能）
# 通常移動で進める距離: 1 * max(0, t - K)

# 距離Gを移動するのに必要な最短時間を求める
# 全力ダッシュをk秒（k <= K）使うとする
# 全力ダッシュでD*k進み、残りG - D*kを通常速度1で進む
# 合計時間 = k + (G - D*k) = G - k*(D-1)

# もし全力ダッシュだけで到着できる場合（D*K >= G）
# 必要時間 = G / D （全力ダッシュのみ）
# より正確には、全力ダッシュをt秒すると距離D*tを移動
# G <= D*K なら、必要時間は G/D

if D * K >= G:
    # 全力ダッシュだけで到着可能
    takahashi_time = G / D
else:
    # K秒全力ダッシュ + 残りを通常移動
    # 全力ダッシュでD*K進み、残りG - D*Kを速度1で進む
    takahashi_time = K + (G - D * K)

if takahashi_time <= train_time:
    print("Yes")
else:
    print("No")

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