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
", kase, d, ans);
  }
  return 0;
}