SGU144 Meeting
题目大意
两个人约会见面,他们可能在X点到Y点之间的任意时刻到达
先到的人会等另一个人Z分钟,问两人能够见面的概率
算法思路
几何概型,答案等于合法区域中 | X - Y | <= Z 表示区域的面积,除以合法区域的面积
时间复杂度: O(1)
代码
/**
* Copyright © 2015 Authors. All rights reserved.
*
* FileName: 144.cpp
* Author: Beiyu Li <sysulby@gmail.com>
* Date: 2015-06-18
*/
#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
typedef long long LL;
typedef pair<int, int> Pii;
const int inf = 0x3f3f3f3f;
const LL infLL = 0x3f3f3f3f3f3f3f3fLL;
int main()
{
int x, y;
double z;
scanf("%d%d%lf", &x, &y, &z);
double l = 60 * (y - x);
double p = (l - z) / l;
printf("%.7f\n", 1 - p * p);
return 0;
}