HDU 1155 Bungee Jumping

题意：英语水平太次…………读了好久好久好久才读懂OTZ

James Bond要逃跑，跑到一个桥边上，要跳到地面，桥边有个有弹性的绳子长度为l，如果他跳下去能到达地面，但速度超过10就会摔死，否则能成功降落，如果不能到达地面则被吊在绳子上（吐槽：为什么不看接近地面就直接跳下去= =）。桥的高度为s，人的重力为w × 9.81，绳的弹力为k × Δl。

解法：一个物理题……物理太渣算了好久……首先看绳长是不是比桥的高度长，如果绳更长则这个人一定会接触地，则在他接触地的时候没有弹力的作用，列出式子：1 / 2mv ^ 2 = mgh，解出速度，如果速度大于10则摔死，否则安全落地。

如果绳没有桥长，则算一下绳最长能抻多长，此时速度为0，重力势能等于弹性势能：mgh = 1 / 2k × l' ^ 2，解出绳最长长度l'，如果l'没有桥长则这人被吊在绳上，否则算一下接触桥的时候的速度，大于10摔死，小于10存活，重力势能等于弹性势能加动能：mgh = 1 / 2mv ^ 2 + 1 / 2k × (h - l) ^ 2，解v。

代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<math.h>
#include<limits.h>
#include<time.h>
#include<stdlib.h>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#define LL long long
using namespace std;
const double g = 9.81;
int main()
{double k, l, s, w;while(cin >> k >> l >> s >> w && !(k == 0 && l == 0 && s == 0 && w == 0)){if(l >= s){double v = sqrt(2 * g * s);if(v > 10)puts("Killed by the impact.");elseputs("James Bond survives.");continue;}double s1;s1 = sqrt(2 * w * g * s / k);if(s1 + l >= s){double v;v = sqrt(2 * g * s - k * (s - l) * (s - l) / w);if(v > 10)puts("Killed by the impact.");elseputs("James Bond survives.");}elseputs("Stuck in the air.");}return 0;
}

转载于:https://www.cnblogs.com/Apro/p/4433153.html

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