LightOJ 1311 Unlucky Bird (物理题)

题意：有点长，意思是说有一个鸟，在两列火车之间不停的来回飞，两列相距为d时，都开始减速，直到最后停止下来，正好是相距0米，

现在给定两列车的速度和减速时的加速度，和鸟的速度求 d 和鸟飞过的路程。

析：就是一个简单的追及相遇问题，注意的是求的飞行时间时，要计算两列火车制动时间最长的那个。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#define debug() puts("++++");

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e16;

const double inf = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 1e5 + 10;

const int mod = 1e9 + 7;

const int dr[] = {-1, 0, 1, 0};

const int dc[] = {0, 1, 0, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c){

  return r >= 0 && r < n && c >= 0 && c < m;

}

int main(){

  int T;  cin >> T;

  for(int kase = 1; kase <= T; ++kase){

    double v1, v2, v3, a1, a2;

    cin >> v1 >> v2 >> v3 >> a1 >> a2;

    double d = v1 * v1 / (2.0*a1) + v2 * v2 / (2.0*a2);

    double ans = v3 * max(v1 / a1, v2 / a2);

    printf("Case %d: %.10f %.10f\n", kase, d, ans);

  }

  return 0;

}

巴特西

LightOJ 1311 Unlucky Bird (物理题)

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