Han Move(细节题)

Problem 1609 - Han Move

Time Limit: 1000MS   Memory Limit: 65536KB    Total Submit: 620  Accepted: 162  Special Judge: No

Description

Cyy and Fzz are Han Move lovers. One day, they gather together to run. They choose a circular track whose length is L m. Cyy’s speed is A m/s and Fzz’s speed is B m/s. They may choose the direction separately, i.e. they may start with the same direction or different direction. As they’re crazy, they’ll run infinitely. Gatevin, their friend, wonders the possibility of their distance is less than D m. The distance is defined as the distance on the track. The possibility is defined as the ratio between the sum of time satisfying the condition and the total time.

Input

The input file consists of multiple test cases ( around 1000000 ). Each test case consists of 5 integers L, A, B, D, Dir in a line. The meanings of L, A, B, D are as described above. Dir means whether they are in the same direction. Dir = 1 means they are in the same direction, while Dir = 0 means they are in the opposite direction. ( 1 <= L, A, B, D <= 32768, 0 <= Dir <= 1 )

Output

For each test case, output the possibility rounded to 6 demical places in a line.

Sample Input

1200 200 400 300 0 1200 200 400 300 1

Sample Output

0.500000 0.500000

题解：两个人正反跑无限跑，问两个人距离为D以内的概率；注意细节；

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int INF = 0x3f3f3f3f;
typedef long long LL;
int main(){
    double L,A,B,D,ans;
    int Dir;
    while(~scanf("%lf%lf%lf%lf%d",&L,&A,&B,&D,&Dir)){
        if(D >= L){
            puts("1.000000");
            continue;
        }
        if(D == 0 && A == B && Dir){
            puts("1.000000");
            continue;
        }
        if(D == 0){
            puts("0.000000");
            continue;
        }
        if(Dir){
            if(A == B){
                puts("0.000000");
                continue;
            }
            double p1 = L / fabs(A - B);
            double p2 = 2 * D / fabs(A - B);
            ans = p2 / p1;
        }
        else{
            double p1 = L / fabs(A + B);
            double p2 = 2 * D / fabs(A + B);
            ans = p2 / p1;
        }
        if(ans >= 1){
            ans = 1;
        }
        printf("%.6lf\n",ans);
    }
    return 0;
}