如果对于集合X 的每个元素，该集合不包含元素X + 1 ，则称一个有序整数集是一个特殊集。给定一个整数N ，任务是找到最大元素不大于N的特殊集合的数量。由于特殊集的数量可能非常大，请打印答案模10 9 + 7 。
例子：
Input: N = 3
Output: 5
{1}, {2}, {3}, {1, 3} and {3, 1} are the
only special sets possible.
Input: N = 4
Output: 10
方法：这个问题可以用动态规划解决。创建一个数组dp[][] ，其中dp[i][j]存储以j结尾的长度为i的特殊集合的数量。现在，递推关系将是：
dp[i][j] = dp[i – 1][1] + dp[i – 1][2] + … + dp[i – 1][j – 2]
dp[i][j] can be computed in O(1) by taking the prefix sum of the previous dp[i – 1] once.
现在可以通过将dp[i][n]与factorial(i)相乘来计算大小i的总特殊集。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
typedef long long ll;
const int MAX = 2 * 1000 + 10;
const int MOD = 1e9 + 7;
// To store the states of the dp
ll dp[MAX][MAX];
// Function to return (a + b) % MOD
ll sum(ll a, ll b)
{
return ((a % MOD) + (b % MOD)) % MOD;
}
// Function to return (a * b) % MOD
ll mul(ll a, ll b)
{
return ((a % MOD) * (b % MOD)) % MOD;
}
// Function to return the count
// of special sets
int cntSpecialSet(int n)
{
// Fill the dp[][] array with the answer
// for the special sets of size 1
for (int i = 1; i <= n; i++) {
dp[1][i] = 1;
// Take prefix sum of the current row which
// will be used to fill the next row
dp[1][i] += dp[1][i - 1];
}
// Fill the rest of the dp[][] array
for (int i = 2; i <= n; i++) {
// Recurrence relation
for (int j = 2; j <= n; j++) {
dp[i][j] = dp[i - 1][j - 2];
}
// Calculate the prefix sum
for (int j = 1; j <= n; j++) {
dp[i][j] = sum(dp[i][j], dp[i][j - 1]);
}
}
ll ways(1), ans(0);
for (int i = 1; i <= n; i++) {
// To find special set of size i
ways = mul(ways, i);
// Addition of special sets of all sizes
ans = sum(ans, mul(ways, dp[i][n]));
}
return ans;
}
// Driver code
int main()
{
int n = 3;
cout << cntSpecialSet(n);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
static int MAX = 2 * 1000 + 10;
static int MOD = (int) (1e9 + 7);
// To store the states of the dp
static long [][]dp = new long[MAX][MAX];
// Function to return (a + b) % MOD
static long sum(long a, long b)
{
return ((a % MOD) + (b % MOD)) % MOD;
}
// Function to return (a * b) % MOD
static long mul(long a, long b)
{
return ((a % MOD) * (b % MOD)) % MOD;
}
// Function to return the count
// of special sets
static long cntSpecialSet(int n)
{
// Fill the dp[][] array with the answer
// for the special sets of size 1
for (int i = 1; i <= n; i++)
{
dp[1][i] = 1;
// Take prefix sum of the current row which
// will be used to fill the next row
dp[1][i] += dp[1][i - 1];
}
// Fill the rest of the dp[][] array
for (int i = 2; i <= n; i++)
{
// Recurrence relation
for (int j = 2; j <= n; j++)
{
dp[i][j] = dp[i - 1][j - 2];
}
// Calculate the prefix sum
for (int j = 1; j <= n; j++)
{
dp[i][j] = sum(dp[i][j], dp[i][j - 1]);
}
}
long ways = 1, ans = 0;
for (int i = 1; i <= n; i++)
{
// To find special set of size i
ways = mul(ways, i);
// Addition of special sets of all sizes
ans = sum(ans, mul(ways, dp[i][n]));
}
return ans;
}
// Driver code
public static void main(String[] args)
{
int n = 3;
System.out.println(cntSpecialSet(n));
}
}
// This code is contributed by PrinciRaj1992Python3
# Python3 implementation of the approach
# Function to print the nodes having
# maximum and minimum degree
def minMax(edges, leng, n) :
# Map to store the degrees of every node
m = {};
for i in range(leng) :
m[edges[i][0]] = 0;
m[edges[i][1]] = 0;
for i in range(leng) :
# Storing the degree for each node
m[edges[i][0]] += 1;
m[edges[i][1]] += 1;
# maxi and mini variables to store
# the maximum and minimum degree
maxi = 0;
mini = n;
for i in range(1, n + 1) :
maxi = max(maxi, m[i]);
mini = min(mini, m[i]);
# Printing all the nodes
# with maximum degree
print("Nodes with maximum degree : ",
end = "")
for i in range(1, n + 1) :
if (m[i] == maxi) :
print(i, end = " ");
print()
# Printing all the nodes
# with minimum degree
print("Nodes with minimum degree : ",
end = "")
for i in range(1, n + 1) :
if (m[i] == mini) :
print(i, end = " ");
# Driver code
if __name__ == "__main__" :
# Count of nodes and edges
n = 4; m = 6;
# The edge list
edges = [[ 1, 2 ], [ 1, 3 ],
[ 1, 4 ], [ 2, 3 ],
[ 2, 4 ], [ 3, 4 ]];
minMax(edges, m, 4);
# This code is contributed by AnkitRai01C#
// C# implementation of the approach
using System;
class GFG
{
static int MAX = 2 * 1000 + 10;
static int MOD = (int) (1e9 + 7);
// To store the states of the dp
static long [,]dp = new long[MAX, MAX];
// Function to return (a + b) % MOD
static long sum(long a, long b)
{
return ((a % MOD) + (b % MOD)) % MOD;
}
// Function to return (a * b) % MOD
static long mul(long a, long b)
{
return ((a % MOD) * (b % MOD)) % MOD;
}
// Function to return the count
// of special sets
static long cntSpecialSet(int n)
{
// Fill the dp[,] array with the answer
// for the special sets of size 1
for (int i = 1; i <= n; i++)
{
dp[1, i] = 1;
// Take prefix sum of the current row which
// will be used to fill the next row
dp[1, i] += dp[1, i - 1];
}
// Fill the rest of the dp[,] array
for (int i = 2; i <= n; i++)
{
// Recurrence relation
for (int j = 2; j <= n; j++)
{
dp[i, j] = dp[i - 1, j - 2];
}
// Calculate the prefix sum
for (int j = 1; j <= n; j++)
{
dp[i, j] = sum(dp[i, j], dp[i, j - 1]);
}
}
long ways = 1, ans = 0;
for (int i = 1; i <= n; i++)
{
// To find special set of size i
ways = mul(ways, i);
// Addition of special sets of all sizes
ans = sum(ans, mul(ways, dp[i, n]));
}
return ans;
}
// Driver code
public static void Main(String[] args)
{
int n = 3;
Console.WriteLine(cntSpecialSet(n));
}
}
// This code is contributed by Princi Singh输出：
5
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