SGU 144. Meeting（概率）

Two of the three members of the winning team of one of the ACM regional contests are going to meet in order to train for the upcoming World Finals. They decided that they will meet sometime between X o'clock and Y o'clock. Because they never get anywhere on time (they were late even on the day of the regional contest), they did not set an exact time when they will meet. However, they decided that the one who gets first at the meeting point will not wait more than Z minutes for the other one (they calculated that, if the other one will not come within Zminutes from the arrival of the first of them, then it is very probable that he will not show up at all). 
Knowing that, in the end, both of them will show up at some time between X o'clock and Y o'clock (not necessarily after an integer number of minutes), compute which is the probability that they will actually meet.

Input

The input will contain 2 integer numbers X and Y (0<=X<Y<=24) and one real number Z ( 0 < Z <= 60*(Y-X) ).

Output

You should output the required probability with 7 decimal digits (rounded according to the 8th decimal digit).

Sample Input

11 12 20.0

Sample Output

0.5555556

 

题意：两个人要见面，他两会在[x,y]这个区间内到达目的地，先到的一个人最多会等第二个人z分钟，问他两能在[x,y]区间相遇的概率是多少。

设l = (y-x)*60，那么要求的概率就如下图所示蓝色区域的面积占总面积的比例。

#include<vector>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#define LL long longusing namespace std;int main(void){    int x,y;    double z;    scanf("%d%d",&x,&y);    scanf("%lf",&z);    double l = (y-x)*60;    double ans = 1.0 - (l-z)*(l-z)/(l*l);    printf("%.7f\n",ans);    return 0;}