最多有W个小门的B球得分R的得分方式

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📜 最多有W个小门的B球得分R的得分方式

📅 最后修改于: 2021-06-25 12:15:34 🧑 作者: Mango

给定三个整数R ， B和W ，分别表示奔跑，球和小门的数目。在板球比赛中，单个球可以得分0、1、2、3、4、6或小门。任务是计算团队最多得分W的B球准确得分R的方式。由于方法的数量很大，因此请打印答案模1000000007 。

例子：

Input: R = 4, B = 2, W = 2
Output: 7
The 7 ways are:
0, 4
4, 0
Wicket, 4
4, Wicket
1, 3
3, 1
2, 2

Input: R = 40, B = 10, W = 4
Output: 653263

为什么编程需要懂一点英语

方法：可以使用动态编程和组合技术解决问题。重复发生将有6种状态，我们首先从运行= 0 ，球= 0和检票口= 0开始。这些状态将是：

如果一个团队在一个球中得分为1，则奔跑=奔跑+ 1 ，球=球+ 1 。
如果一支球队的得分是2，那么球=球+ 2 ，球=球+ 1 。
如果一支球队的得分是3 ，则球=球+ 3 ，球=球+ 1 。
如果一支球队得分为4，那么一个球=奔跑=奔跑+ 4 ，球=球+ 1 。
如果一支球队得分为6 ，则球=球+ 6 ，球=球+ 1 。
如果一个团队没有得分，则没有球，则奔跑=奔跑，球=球+ 1 。
如果一个团队在一个球上丢掉1个小门，那么奔跑=奔跑，小球=球+ 1 ，小门=小门+ 1 。

DP将包含三个状态，运行状态最多为6 * Balls ，因为它是最大可能的状态。因此， dp [i] [j] [k]表示在失去k个小门的情况下，可以准确地在j个球中对i进行得分的方式数。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
#define mod 1000000007
#define RUNMAX 300
#define BALLMAX 50
#define WICKETMAX 10
  
// Function to return the number of ways
// to score R runs in B balls with
// at most W wickets
int CountWays(int r, int b, int l, int R, int B, int W,
              int dp[RUNMAX][BALLMAX][WICKETMAX])
{
  
    // If the wickets lost are more
    if (l > W)
        return 0;
  
    // If runs scored are more
    if (r > R)
        return 0;
  
    // If condition is met
    if (b == B && r == R)
        return 1;
  
    // If no run got scored
    if (b == B)
        return 0;
  
    // Already visited state
    if (dp[r][b][l] != -1)
        return dp[r][b][l];
  
    int ans = 0;
  
    // If scored 0 run
    ans += CountWays(r, b + 1, l, R, B, W, dp);
    ans = ans % mod;
  
    // If scored 1 run
    ans += CountWays(r + 1, b + 1, l, R, B, W, dp);
    ans = ans % mod;
  
    // If scored 2 runs
    ans += CountWays(r + 2, b + 1, l, R, B, W, dp);
    ans = ans % mod;
  
    // If scored 3 runs
    ans += CountWays(r + 3, b + 1, l, R, B, W, dp);
    ans = ans % mod;
  
    // If scored 4 runs
    ans += CountWays(r + 4, b + 1, l, R, B, W, dp);
    ans = ans % mod;
  
    // If scored 6 runs
    ans += CountWays(r + 6, b + 1, l, R, B, W, dp);
    ans = ans % mod;
  
    // If scored no run and lost a wicket
    ans += CountWays(r, b + 1, l + 1, R, B, W, dp);
    ans = ans % mod;
  
    // Memoize and return
    return dp[r][b][l] = ans;
}
  
// Driver code
int main()
{
    int R = 40, B = 10, W = 4;
  
    int dp[RUNMAX][BALLMAX][WICKETMAX];
    memset(dp, -1, sizeof dp);
  
    cout << CountWays(0, 0, 0, R, B, W, dp);
  
    return 0;
}

Java

// Java implementation of the approach
  
class GFG
{
       
static int mod = 1000000007;
static int RUNMAX = 300;
static int BALLMAX = 50;
static int WICKETMAX = 10;
   
// Function to return the number of ways
// to score R runs in B balls with
// at most W wickets
static int CountWays(int r, int b, int l,
                    int R, int B, int W,
                            int [][][]dp)
{
   
    // If the wickets lost are more
    if (l > W)
        return 0;
   
    // If runs scored are more
    if (r > R)
        return 0;
   
    // If condition is met
    if (b == B && r == R)
        return 1;
   
    // If no run got scored
    if (b == B)
        return 0;
   
    // Already visited state
    if (dp[r][b][l] != -1)
        return dp[r][b][l];
   
    int ans = 0;
   
    // If scored 0 run
    ans += CountWays(r, b + 1, l, R, B, W, dp);
    ans = ans % mod;
   
    // If scored 1 run
    ans += CountWays(r + 1, b + 1, l, R, B, W, dp);
    ans = ans % mod;
   
    // If scored 2 runs
    ans += CountWays(r + 2, b + 1, l, R, B, W, dp);
    ans = ans % mod;
   
    // If scored 3 runs
    ans += CountWays(r + 3, b + 1, l, R, B, W, dp);
    ans = ans % mod;
   
    // If scored 4 runs
    ans += CountWays(r + 4, b + 1, l, R, B, W, dp);
    ans = ans % mod;
   
    // If scored 6 runs
    ans += CountWays(r + 6, b + 1, l, R, B, W, dp);
    ans = ans % mod;
   
    // If scored no run and lost a wicket
    ans += CountWays(r, b + 1, l + 1, R, B, W, dp);
    ans = ans % mod;
   
    // Memoize and return
    return dp[r][b][l] = ans;
}
   
// Driver code
public static void main(String[] args)
{
    int R = 40, B = 10, W = 4;
   
    int[][][] dp = new int[RUNMAX][BALLMAX][WICKETMAX];
    for(int i = 0; i < RUNMAX;i++)
        for(int j = 0; j < BALLMAX; j++)
            for(int k = 0; k < WICKETMAX; k++)
    dp[i][j][k]=-1;
   
    System.out.println(CountWays(0, 0, 0, R, B, W, dp));
}
}
  
// This code has been contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
mod = 1000000007
RUNMAX = 300
BALLMAX = 50
WICKETMAX = 10
  
# Function to return the number of ways
# to score R runs in B balls with
# at most W wickets
def CountWays(r, b, l, R, B, W, dp):
      
    # If the wickets lost are more
    if (l > W):
        return 0;
  
    # If runs scored are more
    if (r > R):
        return 0;
  
    # If condition is met
    if (b == B and r == R):
        return 1;
  
    # If no run got scored
    if (b == B):
        return 0;
  
    # Already visited state
    if (dp[r][b][l] != -1):
        return dp[r][b][l]
          
    ans = 0;
  
    # If scored 0 run
    ans += CountWays(r, b + 1, l,
                     R, B, W, dp);
    ans = ans % mod;
  
    # If scored 1 run
    ans += CountWays(r + 1, b + 1, l, 
                     R, B, W, dp);
    ans = ans % mod;
  
    # If scored 2 runs
    ans += CountWays(r + 2, b + 1, l,
                     R, B, W, dp);
    ans = ans % mod;
  
    # If scored 3 runs
    ans += CountWays(r + 3, b + 1, l,
                     R, B, W, dp);
    ans = ans % mod;
  
    # If scored 4 runs
    ans += CountWays(r + 4, b + 1, l, 
                     R, B, W, dp);
    ans = ans % mod;
  
    # If scored 6 runs
    ans += CountWays(r + 6, b + 1, l,
                     R, B, W, dp);
    ans = ans % mod;
  
    # If scored no run and lost a wicket
    ans += CountWays(r, b + 1, l + 1, 
                     R, B, W, dp);
    ans = ans % mod;
      
    # Memoize and return
    dp[r][b][l] = ans
      
    return ans;
      
# Driver code    
if __name__=="__main__":
      
    R = 40
    B = 10
    W = 40
      
    dp = [[[-1 for k in range(WICKETMAX)] 
               for j in range(BALLMAX)] 
               for i in range(RUNMAX)]
      
    print(CountWays(0, 0, 0, R, B, W, dp))
  
# This code is contributed by rutvik_56

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
using System.Linq;
  
class GFG
{
      
static int mod = 1000000007;
static int RUNMAX = 300;
static int BALLMAX = 50;
static int WICKETMAX = 10;
  
// Function to return the number of ways
// to score R runs in B balls with
// at most W wickets
static int CountWays(int r, int b, int l,
                    int R, int B, int W,
                            int [,,]dp)
{
  
    // If the wickets lost are more
    if (l > W)
        return 0;
  
    // If runs scored are more
    if (r > R)
        return 0;
  
    // If condition is met
    if (b == B && r == R)
        return 1;
  
    // If no run got scored
    if (b == B)
        return 0;
  
    // Already visited state
    if (dp[r, b, l] != -1)
        return dp[r, b, l];
  
    int ans = 0;
  
    // If scored 0 run
    ans += CountWays(r, b + 1, l, R, B, W, dp);
    ans = ans % mod;
  
    // If scored 1 run
    ans += CountWays(r + 1, b + 1, l, R, B, W, dp);
    ans = ans % mod;
  
    // If scored 2 runs
    ans += CountWays(r + 2, b + 1, l, R, B, W, dp);
    ans = ans % mod;
  
    // If scored 3 runs
    ans += CountWays(r + 3, b + 1, l, R, B, W, dp);
    ans = ans % mod;
  
    // If scored 4 runs
    ans += CountWays(r + 4, b + 1, l, R, B, W, dp);
    ans = ans % mod;
  
    // If scored 6 runs
    ans += CountWays(r + 6, b + 1, l, R, B, W, dp);
    ans = ans % mod;
  
    // If scored no run and lost a wicket
    ans += CountWays(r, b + 1, l + 1, R, B, W, dp);
    ans = ans % mod;
  
    // Memoize and return
    return dp[r, b, l] = ans;
}
  
// Driver code
static void Main()
{
    int R = 40, B = 10, W = 4;
  
    int[,,] dp = new int[RUNMAX, BALLMAX, WICKETMAX];
    for(int i = 0; i < RUNMAX;i++)
        for(int j = 0; j < BALLMAX; j++)
            for(int k = 0; k < WICKETMAX; k++)
    dp[i, j, k]=-1;
  
    Console.WriteLine(CountWays(0, 0, 0, R, B, W, dp));
}
}
  
// This code is contributed by mits

输出：

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