给定一个表示骰子数量的整数N ,任务是找到通过将N 个骰子放在一起可以获得的每个可能值的概率。
例子:
Input: N = 1
Output:
1: 0.17
2: 0.17
3: 0.17
4: 0.17
5: 0.17
6: 0.17
Explanation: On throwing a dice, the probability of all values from [1, 6] to appear at the top is 1/6 = 0.17
Input: N = 2
Output:
2: 0.028
3: 0.056
4: 0.083
5: 0.11
6: 0.14
7: 0.17
8: 0.14
9: 0.11
10: 0.083
11: 0.056
12: 0.028
Explanation: The possible values of the sum of the two numbers that appear at the top on throwing two dices together ranges between [2, 12].
方法:思路是使用动态规划和DP表来存储每个可能值的概率。
- 存储在投掷 1 个骰子时可能出现的所有 6 个数字的概率。
- 现在,对于 N=2,[2, 12] 之间所有可能总和的概率等于两个数字相加后的相应概率的乘积之和。例如,
Probability of 4 on throwing 2 dices = (Probability of 1 ) * ( Probability of 3) + (Probability of 2) * ( Probability of 2) + (Probability of 3 ) * ( Probability of 1)
- 因此对于 N 个骰子,
Probability of Sum S = (Probability of 1) * (Probability of S – 1 using N -1 dices) + (Probability of 2) * (Probability of S – 2 using N-1 dices) + ….. + (Probability of 6) * (Probability of S – 6 using N -1 dices)
- 因此,为了解决这个问题,我们需要使用自顶向下的方法将 dp[][] 表从 2 填充到 N:
dp[i][x] = dp[1][y] + dp[i-1][z] where x = y + z and i denotes the number of dices
- 显示为 N 存储的所有概率作为答案。
下面是上述方法的实现:
C++
// C++ Program to calculate
// the probability of
// all the possible values
// that can be obtained
// throwing N dices
#include
using namespace std;
void dicesSum(int n)
{
// Store the probabilities
vector
Java
// Java program to calculate
// the probability of all the
// possible values that can
// be obtained throwing N dices
import java.io.*;
import java.util.*;
class GFG{
static void dicesSum(int n)
{
// Store the probabilities
double[][] dp = new double[n + 1][6 * n + 1];
// Precompute the probabilities
// for values possible using 1 dice
for(int i = 1; i <= 6; i++)
dp[1][i] = 1 / 6.0;
// Compute the probabilies
// for all values from 2 to N
for(int i = 2; i <= n; i++)
for(int j = i - 1; j <= 6 * (i - 1); j++)
for(int k = 1; k <= 6; k++)
{
dp[i][j + k] += (dp[i - 1][j] *
dp[1][k]);
}
// Print the result
for(int i = n; i <= 6 * n; i++)
{
System.out.println(i + " " +
Math.round(dp[n][i] * 1000.0) /
1000.0);
}
}
// Driver Code
public static void main(String[] args)
{
int n = 2;
dicesSum(n);
}
}
// This code is contributed by jithin
Python3
# Python3 program to calculate
# the probability of all the
# possible values that can
# be obtained throwing N dices
def diceSum(n):
# Inititalize a 2d array upto
# (n*total sum possible) sum
# with value 0
dp = [[ 0 for j in range(n * 6)]
for i in range(n + 1)]
# Store the probability in a
# single throw for 1,2,3,4,5,6
for i in range(6):
dp[1][i] = 1 / 6
# Compute the probabilies
# for all values from 2 to N
for i in range(2, n + 1):
for j in range(len(dp[i - 1])):
for k in range(6):
if (dp[i - 1][j] != 0 and
dp[i - 1][k] != 0):
dp[i][j + k] += (dp[i - 1][j] *
dp[1][k])
# Print the result
for i in range(len(dp[n]) - n + 1):
print("%d %0.3f" % (i + n, dp[n][i]))
# Driver code
n = 2
# Call the function
diceSum(n)
# This code is contributed by dipesh99kumar
C#
// C# program to calculate
// the probability of all the
// possible values that can
// be obtained throwing N dices
using System;
class GFG {
static void dicesSum(int n)
{
// Store the probabilities
double[,] dp = new double[n + 1,6 * n + 1];
// Precompute the probabilities
// for values possible using 1 dice
for(int i = 1; i <= 6; i++)
dp[1,i] = 1 / 6.0;
// Compute the probabilies
// for all values from 2 to N
for(int i = 2; i <= n; i++)
for(int j = i - 1; j <= 6 * (i - 1); j++)
for(int k = 1; k <= 6; k++)
{
dp[i,j + k] += (dp[i - 1,j] *
dp[1,k]);
}
// Print the result
for(int i = n; i <= 6 * n; i++)
{
Console.WriteLine(i + " " +
Math.Round(dp[n,i] * 1000.0) /
1000.0);
}
}
static void Main() {
int n = 2;
dicesSum(n);
}
}
// This code is contributed by divyesh072019
Javascript
2 0.028
3 0.056
4 0.083
5 0.11
6 0.14
7 0.17
8 0.14
9 0.11
10 0.083
11 0.056
12 0.028
时间复杂度: O(N 2 )
辅助空间: O(N 2 )