Code shoe set - mt2095 · zigzag jump

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Zigzag path

• Zigzag path .

The time limit ：1 second
Space restriction ：128M

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Title Description

Small brother is very cruel , They often send their teammates to some strange places . This time, , He put the zigzag path on jiejie At the foot of . because jiejie Don't want to be disturbed by your rhythm , So he chose to escape from heaven .jiejie Playing with the prince , Skill eq Continuous moves can displace x Yards away ,eq Displacement does not consume time . meanwhile ,jiejie The movement speed of is n Yards per second , The radius of the zigzag jump is r code .jiejie Your skills may be in cd,e and q Any skill in cd Can't use skill displacement . The effective time of zigzag jump is 2 second . Now he wants to know if he can escape the winding path
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Input description

5 It's an integer

e：e Skill cd（ by 0 Can be used when ）

q：q Skill cd

x： Moving distance

n： Velocity of movement

r： radius

Data range

e、q Is a nonnegative integer ,x、r、n As a positive integer

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Output description

If you can escape the output yes

Otherwise output no

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Example 1

Input

0 0 500 100 700

Output

yes

I have read this question for a long time _{（ Maybe it's because I haven't played much LOL）}

After reading the topic, it's very simple

The main idea of the topic

Ask you $2$ Seconds , Can I move $\ge r$ Distance of .

There are two ways to move ：

1. Walk ： It can move every second $n$ rice
2. teleport ： It doesn't take time , Can move instantly $x$ rice . But the premise is $e\le 2Andq\le 2$

Topic analysis

After reading the topic , It can be simulated directly .

Two seconds , Walking can move $2\times n$ rice ,

If $e\le 2Andq\le 2$ , Then you can blink again $x$ rice .

See whether the final moving distance $\ge r$ that will do .

AC Code

/* * @Author: LetMeFly * @Date: 2022-07-21 10:29:45 * @LastEditors: LetMeFly * @LastEditTime: 2022-07-21 10:33:26 */
#include <bits/stdc++.h>
using namespace std;
#define mem(a) memset(a, 0, sizeof(a))
#define dbg(x) cout << #x << " = " << x << endl
#define fi(i, l, r) for (int i = l; i < r; i++)
#define cd(a) scanf("%d", &a)
typedef long long ll;
int main() {
    
    int e, q, x, n, r;
    cin >> e >> q >> x >> n >> r;
    r -= 2 * n;
    if (e <= 2 && q <= 2)
        r -= x;
    if (r <= 0)
        puts("yes");
    else
        puts("no");
    return 0;
}

Although the code can be copied , But it's better to knock after understanding

Originality is not easy. , Reprint please attach Link to the original text Oh ~
Tisfy：https://letmefly.blog.csdn.net/article/details/125909920