程式道路，必為豐富

內容

有一群人，輪流骰骰子(不一定是六面骰)，算出指定的人贏的機率。

Input
Input will contain an integer S (S ≤ 1000) at first, which indicates how many sets of inputs are there.
The next S lines will contain S sets of inputs. Each line contain an integer N (N ≤ 1000) which denotes
the number players, a floating point number p which indicates the probability of the happening of a
successful event in a single throw (If success means getting 3 then p is the probability of getting 3 in
a single throw. For a normal dice the probability of getting 3 is 1/6), and I (I ≤ N) the serial of the
player whose winning probability is to be determined (Serial no varies from 1 to N). You can assume
that no invalid probability (p) value will be given as input.
第一行S(S ≤ 1000)
接下來S行有3三個數字
N(N ≤ 1000) p I(I ≤ N) 
N是總參加人數
p是該骰子贏的機率
I是指定的人

Output
For each set of input, output in a single line the probability of the I-th player to win. The output
floating point number will always have four digits after the decimal point as shown in the sample
output.
每一行輸出皆為小數點後4位

Sample Input
2
2 0.166666 1
2 0.166666 2

Sample Output
0.5455
0.4545

C++

#include<iostream>
#include <cmath>
#include <iomanip>
using namespace std;  
int s;  
int main() 
{
          
	while(cin>>s)//測資數量
	{
		int n,i;
		long double p;
		while(s--)
		{             
			cin>>n>>p>>i;//輸入數字 
			int round=0;
			long double nowp=1,ans=0;//nowp是當前回合的贏機率 
			do
			{
				 
				nowp=pow((1-p),round*n)*pow((1-p),i-1)*p;//pow((1-p),round*n) 是玩到第round+1輪的機率 
														 //pow((1-p),i-1)是該輪前面的人輸的機率
														 //p 是自己勝率的機率 
				ans+=nowp;//把每回合的贏的機率相加就是答案 
				round++;
				
			}while(nowp>0.00000000000001);//0多打幾個，才算的準 
			
			cout<<fixed<<setprecision(4)<<ans<<endl;;
		}
		
	}
	return 0;
}
 

python

from decimal import getcontext, Decimal
import decimal
s = int(input())
for i in range(s):
    n = input()
    n = n.split()
    p = float(n[1])
    round = 0
    nowp = 0.000000
    ans = 0.000
    nowp = ((1-p)**(int(n[2])-1))*p
    while nowp > 0.0000000000001:
        ans += nowp
        round += 1
        nowp = ((1-p)**(round*int(n[0])))*((1-p)**(int(n[2])-1))*p

    ans = Decimal(ans).quantize(Decimal('0.0000'))
    print(ans)