1 2 3 1 2 4 1 2 5 1 3 4 1 3 5
4 5 6 3 5 6 3 4 6 2 5 6 2 4 6
//法一（暴力）//本题相当于对123456每一位进行标记，如果0就放到第一行，如果为1就放到第二行，所以0的个数要和1的个//数相等，又因为123456本身就是一个有序的，所以要想保证列上的递增，0的个数始终要大于等于1的个数。#include<stdio.h>#include<math.h>#define n 3int main(){int i,j;int count0,count1;int count = 0;int num;for(i = 0;i < pow(2,2 * n);i ++){num = i;count0 = count1 = 0;for(j = 0;j < 2 * n;j ++){if(num % 2 == 0){count0 ++;}else{count1 ++;}num = num / 2;if(count1 > count0)break;} if(count1 == count0 && count0 == n)count ++;}printf("%dn",count);return 0;}//法二#include<stdio.h>int f(int n){int i;int count = 0;if(n == 1 || n == 0)return 1;for(i = 0;i < n;i ++)count = count + f(i) * f(n - 1 - i);return count;}int main(){int n = 3;printf("%dn",f(n));return 0;}