bnu29064--Coin Water Title II —————— "event probability"

Coin water problem iitime limit:1000msmemory limit:65536kb64-bit integer IO format:%lld Java class name: Main Submit Status pid:29064

Chubby has a coin with a positive and negative face. If you toss this coin, the probability of its facing upward is p, and the probability of the opposite facing up is 1-p. Now, Chubby wants to use this coin to produce equal probability decisions (50% to 50%). Of course, it's not possible to just toss it once. The strategy for chubby is this: every decision needs to be tossed two times, if it is either positive or negative, then make a decision again; if it is a positive one, then if the first is facing upward, then it is said that if the first is a face-up, then it is considered to be the opposite. In this way, a fair decision can be made.

The question now is, given a p, how many times will the fat average be thrown to get a decision (i.e. no more throwing)?

Input

The first line contains an integer N (n<=100) that represents the number of test data.

Next consists of n rows, one test data per line, including a 3-bit floating point P (0<p<1).

Output

For each test data, the output line, including a floating-point number, represents the average count of the small-fat coin toss.

The result retains two decimal places.

Sample Input

30.5000.8000.300

Sample Output

4.006.254.76

#include <bits/stdc++.h>using namespace Std;int main () {    int n;    scanf ("%d", &n);    while (n--) {        double p;        scanf ("%lf", &p);        Double ans=1.0/(1.0-p*p-(1-p) * (1-p));        printf ("%.2lf\n", 2*ans);    }    return 0;}

　　

#include <bits/stdc++.h>using namespace Std;int main () {    int n;    scanf ("%d", &n);    while (n--) {        double p;        scanf ("%lf", &p);        Double ans=1.0/p+1.0/(1.0-p);        printf ("%.2lf\n", ans);    }    return 0;}

　　

bnu29064--Coin Water Title II —————— "event probability"