Acwing---889. 01 sequence that meets the conditions (Java)_mathematics knowledge_combination counting_cattelland number

Original title link

①. Title

②. Thinking

• The problem is directly transformed into a walking grid. If 0 means going to the right and 1 means going up, then the number of 0s in any prefix is ​​not less than 1 and the number is converted to any point on the path, the abscissa is greater than or equal to the vertical coordinate. The number of 0 in all prefixes greater than 1 is below the red line. If the path through the red line is symmetric, then their end points will be mapped to (n−1, n+1), as long as the end point is calculated as (n−1, n+1). 1, n+1) How many paths are there, and then subtracting it from the total is the final answer.

• Indicates the path from (0,0) to (n,n). The green line and below are legal. If you touch the red line, it is illegal.
  

• Algorithm core: every illegal path from (0,0) to (n,n) corresponds to an illegal path from (0,0) to (n-1,n+1)

• Directly deduced the formula Cn2n−Cn−12n, simplified to Cn2n/(n+1)
  

③. Learning points

组合计数_卡特兰数

④. Code implementation

import java.util.Scanner;

public class Main {
    
    
	static int m=(int)1e9+7;
	public static void main(String[] args) {
    
    
		Scanner sc = new Scanner(System.in);
		int n=sc.nextInt();
		long res=1;
		//直接套用推出来的公式
		for (int i = 1,j=2*n; i <=n; i++,j--) {
    
    
			res=res*j%m; //求2n*(2n-1)*(2n-2)*...*(2n-n+1)
			res=res*qmi(i,m-2,m)%m; //快速幂求n的阶乘的逆元
		}
		System.out.println(res*qmi(n+1,m-2,m)%m); //最后乘上n+1的逆元
	}
	
	//快速幂求 a^k%m
	static long qmi(long a,long k,long m) {
    
    
		long res=1;
		while(k!=0) {
    
    
			if((k&1)==1) res=res*a%m;
			a=a*a%m;
			k>>=1;
		}
		return res;
	}	
}