给定一个代表学生人数的数字N ，任务是计算所有可能的方式，根据他们的CGPA /分数对他们进行排名，同时考虑到两个或更多学生可以具有相同的排名。由于答案可能很大，因此请使用10 9 + 7取模。
例子：
Input: N = 1
Output: 1
Explanation:
There is only one way to rank a student, irrespective of his marks.
Input: N = 2
Output: 2
Explanation:
In the following two ways the ranks can be distributed among two students:
The high scoring student can be awarded the first rank and the low scoring student the second rank or both of them can be awarded the same rank if they have equal scores.
方法：这个问题的想法是使用贝尔号码。
- 响铃号是一个数字，它计算一个集合的可能分区。因此，第N个响铃号是一组大小为N的非空子集的数量。
- 例如，让我们考虑N = 3的集合{1、2、3}。与N = 3对应的响铃编号是5。这表示可以将给定集合划分为以下5个非空子集：
{{1}, {2}, {3}}
{{1, 2}, {3}}
{{1, 3}, {2}}
{{2, 3}, {1}}
{{1, 2, 3}}- 显然，上面的铃号表示所有可能的等级。但是，它们不计算子集的排列。
- 因此，通过将每个子集乘以K！(其中K表示各个子集的大小)，我们得到所有可能的排列。
- 由于可以重复相同的子问题，因此我们可以将每个子问题的值存储在数据结构中以优化复杂性。
下面是上述方法的实现：
C++
// C++ program to calculate the number
// of ways to give ranks for N
// students such that same ranks
// are possible
#include
using namespace std;
const int mod = 1e9 + 7;
// Initializing a table in order to
// store the bell triangle
vector > dp;
// Function to calculate the K-th
// bell number
int f(int n, int k)
{
// If we have already calculated
// the bell numbers until the
// required N
if (n < k)
return 0;
// Base case
if (n == k)
return 1;
// First Bell Number
if (k == 1)
return 1;
// If the value of the bell
// triangle has already been
// calculated
if (dp[n][k] != -1)
return dp[n][k];
// Fill the defined dp table
return dp[n][k] = ((k * f(n - 1, k)) % mod
+ (f(n - 1, k - 1)) % mod)
% mod;
}
// Function to return the number
// of ways to give ranks for N
// students such that same ranks
// are possible
long operation(int n)
{
// Resizing the dp table for the
// given value of n
dp.resize(n + 1, vector(n + 1, -1));
// Variables to store the answer
// and the factorial value
long ans = 0, fac = 1;
// Iterating till N
for (int k = 1; k <= n; k++) {
// Simultaneously calculate the k!
fac *= k;
// Computing the K-th bell number
// and multiplying it with K!
ans = (ans + (fac * f(n, k)) % mod)
% mod;
}
return ans;
}
// Driver code
int main()
{
int n = 5;
cout << operation(n) << endl;
return 0;
} Java
// Java program to calculate the number
// of ways to give ranks for N
// students such that same ranks
// are possible
import java.util.*;
class GFG{
static int mod = (int)(1e9 + 7);
// Initializing a table in order to
// store the bell triangle
static int [][]dp;
// Function to calculate the K-th
// bell number
static int f(int n, int k)
{
// If we have already calculated
// the bell numbers until the
// required N
if (n < k)
return 0;
// Base case
if (n == k)
return 1;
// First Bell Number
if (k == 1)
return 1;
// If the value of the bell
// triangle has already been
// calculated
if (dp[n][k] != -1)
return dp[n][k];
// Fill the defined dp table
return dp[n][k] = ((k * f(n - 1, k)) % mod +
(f(n - 1, k - 1)) % mod) % mod;
}
// Function to return the number
// of ways to give ranks for N
// students such that same ranks
// are possible
static long operation(int n)
{
// Resizing the dp table for the
// given value of n
dp = new int[n + 1][n + 1];
for(int i = 0; i < n + 1; i++)
{
for(int j = 0; j < n + 1; j++)
{
dp[i][j] = -1;
}
}
// Variables to store the answer
// and the factorial value
long ans = 0, fac = 1;
// Iterating till N
for(int k = 1; k <= n; k++)
{
// Simultaneously calculate the k!
fac *= k;
// Computing the K-th bell number
// and multiplying it with K!
ans = (ans + (fac * f(n, k)) % mod) % mod;
}
return ans;
}
// Driver code
public static void main(String[] args)
{
int n = 5;
System.out.print(operation(n) + "\n");
}
}
// This code is contributed by amal kumar choubeyPython3
# Python3 program to calculate the number
# of ways to give ranks for N
# students such that same ranks
# are possible
mod = 1e9 + 7
# Initializing a table in order to
# store the bell triangle
dp = [[-1 for x in range(6)]
for y in range(6)]
# Function to calculate the K-th
# bell number
def f(n, k):
# If we have already calculated
# the bell numbers until the
# required N
if (n < k):
return 0
# Base case
if (n == k):
return 1
# First Bell Number
if (k == 1):
return 1
# If the value of the bell
# triangle has already been
# calculated
if (dp[n][k] != -1):
return dp[n][k]
# Fill the defined dp table
dp[n][k] = ((((k * f(n - 1, k)) % mod +
(f(n - 1, k - 1)) % mod) % mod))
return dp[n][k]
# Function to return the number
# of ways to give ranks for N
# students such that same ranks
# are possible
def operation(n):
# Resizing the dp table for the
# given value of n
global dp
# Variables to store the answer
# and the factorial value
ans = 0
fac = 1
# Iterating till N
for k in range(1, n + 1):
# Simultaneously calculate the k!
fac *= k
# Computing the K-th bell number
# and multiplying it with K!
ans = (ans + (fac * f(n, k)) % mod) % mod
return ans
# Driver code
if __name__ == "__main__":
n = 5
print(int(operation(n)))
# This code is contributed by ukaspC#
// C# program to calculate the number
// of ways to give ranks for N
// students such that same ranks
// are possible
using System;
class GFG{
static int mod = (int)(1e9 + 7);
// Initializing a table in order to
// store the bell triangle
static int [,]dp;
// Function to calculate the K-th
// bell number
static int f(int n, int k)
{
// If we have already calculated
// the bell numbers until the
// required N
if (n < k)
return 0;
// Base case
if (n == k)
return 1;
// First Bell Number
if (k == 1)
return 1;
// If the value of the bell
// triangle has already been
// calculated
if (dp[n, k] != -1)
return dp[n, k];
// Fill the defined dp table
return dp[n, k] = ((k * f(n - 1, k)) % mod +
(f(n - 1, k - 1)) % mod) % mod;
}
// Function to return the number
// of ways to give ranks for N
// students such that same ranks
// are possible
static long operation(int n)
{
// Resizing the dp table for the
// given value of n
dp = new int[n + 1, n + 1];
for(int i = 0; i < n + 1; i++)
{
for(int j = 0; j < n + 1; j++)
{
dp[i, j] = -1;
}
}
// Variables to store the answer
// and the factorial value
long ans = 0, fac = 1;
// Iterating till N
for(int k = 1; k <= n; k++)
{
// Simultaneously calculate the k!
fac *= k;
// Computing the K-th bell number
// and multiplying it with K!
ans = (ans + (fac * f(n, k)) % mod) % mod;
}
return ans;
}
// Driver code
public static void Main(String[] args)
{
int n = 5;
Console.Write(operation(n) + "\n");
}
}
// This code is contributed by amal kumar choubey输出：
541