HDU-1115 Bungee Jumping

本文探讨了一个经典的物理问题:詹姆斯·邦德使用弹性绳从不同高度的桥梁跳下时的生存可能性。通过分析弹性绳的参数,包括其劲度系数、原长及邦德的体重等,来判断邦德落地时的状态。

Bungee Jumping

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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Status

Practice

HDU 1155
Description
Once again, James Bond is fleeing from some evil people who want to see him dead. Fortunately, he has left a bungee rope on a nearby highway bridge which he can use to escape from his enemies. His plan is to attach one end of the rope to the bridge, the other end of the rope to his body and jump off the bridge. At the moment he reaches the ground, he will cut the rope, jump into his car and be gone.

Unfortunately, he had not had enough time to calculate whether the bungee rope has the right length, so it is not clear at all what is going to happen when he jumps off the bridge. There are three possible scenarios:
The rope is too short (or too strong), and James Bond will never reach the ground.
The rope is too long (or too weak), and James Bond will be going too fast when he touches the ground. Even for a special agent, this can be very dangerous. You may assume that if he collides at a speed of more than 10 m/s, he will not survive the impact.
The rope’s length and strength are good. James Bond touches the ground at a comfortable speed and can escape.
As his employer, you would like to know whether James Bond survives or whether you should place a job ad for the soon-to-be vacant position in the local newspaper. Your physicists claim that:
The force with which James is pulled towards the earth is
9.81 * w,
where w is his weight in kilograms and 9.81 is the Earth acceleration in meters over squared seconds.
Mr. Bond falls freely until the rope tautens. Then the force with which the bungee rope pulls him back into the sky depends on the current length of the rope and is
k * Δl,
where Δl is the difference between the rope’s current length and its nominal, unexpanded length, and k is a rope-specific constant.
Given the rope’s strength k, the nominal length of the rope l in meters, the height of the bridge s in meters, and James Bond’s body weight w, you have to determine what is going to happen to our hero. For all your calculations, you may assume that James Bond is a point at the end of the rope and the rope has no mass. You may further assume that k, l, s, and w are non-negative and that s < 200.

The input contains several test cases, one test case per line. Each test case consists of four floating-point numbers (k, l, s, and w) that describe the situation. Depending on what is going to happen, your program must print “Stuck in the air.”, “Killed by the impact.”, or “James Bond survives.”. Input is terminated by a line containing four 0s, this line should not be processed.
Input
350 20 30 75
375 20 30 75
400 20 30 75
425 20 30 75
450 20 30 75
400 20 30 50
400 20 30 80
400 20 30 85
0 0 0 0
Output
Killed by the impact.
James Bond survives.
James Bond survives.
James Bond survives.
Stuck in the air.
Stuck in the air.
James Bond survives.
Killed by the impact.
Sample Input
350 20 30 75
375 20 30 75
400 20 30 75
425 20 30 75
450 20 30 75
400 20 30 50
400 20 30 80
400 20 30 85
0 0 0 0
Sample Output
Killed by the impact.
James Bond survives.
James Bond survives.
James Bond survives.
Stuck in the air.
Stuck in the air.
James Bond survives.
Killed by the impact.


数学题:一个从高为S的桥上系这弹性绳跳下来,劲度系数K, 原长L,如果接触地面的速度>10m/s,就会摔死,给定数据,问你这个人跳下来会怎样;
就是分情况讨论就行,我就是相统一所有情况,不小心就忽略了一种;
比赛时竟然没写,都是大长题,看着吓人。


#include<cstdio>
using namespace std;

#define G 9.81
#define eps 1e-8

int main()
{
    double W,K,L,V2,S;
    while(~scanf("%lf %lf %lf %lf",&K,&L,&S,&W),K||W||L||S)
    {
        if(S<L) //这种情况忽略了 
        {
             if(2*G*S>100) printf("Killed by the impact.\n");
             else printf("James Bond survives.\n");
             continue;
        }
        V2=(2*W*G*S-K*(S-L)*(S-L) )/W;
        if(V2<0)
        {
            printf("Stuck in the air.\n");
        }
        else if(V2>100)   printf("Killed by the impact.\n");
        else printf("James Bond survives.\n");
     } 
    return 0;
}
### 关于HDU - 6609 的题目解析 由于当前未提供具体关于 HDU - 6609 题目的详细描述,以下是基于一般算法竞赛题型可能涉及的内容进行推测和解答。 #### 可能的题目背景 假设该题目属于动态规划类问题(类似于多重背包问题),其核心在于优化资源分配或路径选择。此类问题通常会给出一组物品及其属性(如重量、价值等)以及约束条件(如容量限制)。目标是最优地选取某些物品使得满足特定的目标函数[^2]。 #### 动态转移方程设计 如果此题确实是一个变种的背包问题,则可以采用如下状态定义方法: 设 `dp[i][j]` 表示前 i 种物品,在某种条件下达到 j 值时的最大收益或者最小代价。对于每一种新加入考虑范围内的物体 k ,更新规则可能是这样的形式: ```python for i in range(n): for s in range(V, w[k]-1, -1): dp[s] = max(dp[s], dp[s-w[k]] + v[k]) ``` 这里需要注意边界情况处理以及初始化设置合理值来保证计算准确性。 另外还有一种可能性就是它涉及到组合数学方面知识或者是图论最短路等相关知识点。如果是后者的话那么就需要构建相应的邻接表表示图形结构并通过Dijkstra/Bellman-Ford/Floyd-Warshall等经典算法求解两点间距离等问题了[^4]。 最后按照输出格式要求打印结果字符串"Case #X: Y"[^3]。 #### 示例代码片段 下面展示了一个简单的伪代码框架用于解决上述提到类型的DP问题: ```python def solve(): t=int(input()) res=[] cas=1 while(t>0): n,k=list(map(int,input().split())) # Initialize your data structures here ans=find_min_unhappiness() # Implement function find_min_unhappiness() res.append(f'Case #{cas}: {round(ans)}') cas+=1 t-=1 print("\n".join(res)) solve() ```
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