Android解锁图案共有多少种？

九个点

每个点只能连一次 ，如果两个点AB之间有其他点C，在C点被使用之前AB不能直接相连 

两种解法 

一种是直接用简单的9个循环 

#include <stdio.h>

int main(){
	int count=0;
	for(int i1=1; i1<10; i1++){
		for(int i2=1; i2<10; i2++){
			if(i2 == i1) continue;
			if((i2+i1)%2 ==0 ){
				int mid=(i1+i2)/2;
				if(mid == 2 || mid== 8 || mid == 5) continue;
				if(mid == 4 && (i2 == 1 || i2 == 7)) continue;
				if(mid == 6 && (i2 == 3 || i2 == 9)) continue;
			}

			for(int i3=1; i3<10; i3++){
				if(i3 == i1 || i3 == i2) continue;
				if((i2+i3)%2 ==0 ){
					int mid=(i3+i2)/2;
					if((mid == 2 || mid== 8 || mid == 5) && mid != i1) continue;
					if(mid == 4 && i1 != 4 && (i3 == 1 || i3 == 7)) continue;
					if(mid == 6 && i1 != 6 && (i3 == 3 || i3 == 9)) continue;
				}


				for(int i4=1; i4<10; i4++){
					if(i4 == i1 || i4 == i2 || i4 == i3) continue;
					if((i3+i4)%2 ==0 ){
						int mid=(i3+i4)/2;
						if((mid == 2 || mid== 8 || mid == 5) && mid != i1 && mid != i2) continue;
						if(mid == 4 && i1 != 4 && i2!=4 && (i4 == 1 || i4 == 7)) continue;
						if(mid == 6 && i1 != 6 && i2!=6 && (i4 == 3 || i4 == 9)) continue;
					}

					for(int i5=0; i5<10; i5++){
						if(i5 == i1 || i5 == i2 || i5 == i3 || i5 == i4 ) continue;
						if( i5!= 0 && (i4+i5)%2 ==0 ){
							int mid=(i4+i5)/2;
							if((mid == 2 || mid== 8 || mid == 5) && mid != i1 && mid != i2 && mid != i3) continue;
							if(mid == 4 && i1 != 4 && i2!=4 && i3!=4 && (i5 == 1 || i5 == 7)) continue;
							if(mid == 6 && i1 != 6 && i2!=6 && i3!=6 && (i5 == 3 || i5 == 9)) continue;
						}

						if(i5 == 0){
							printf("%d\t%d%d%d%d\n",count+1, i1,i2,i3,i4);
							count++;
						}else{

							for(int i6=0; i6<10; i6++){
								if( i6 == i1 || i6 == i2 || i6 == i3 || i6 == i4 || i6 == i5) continue;
								if( i6 != 0 && (i5+i6)%2 ==0 ){
									int mid=(i5+i6)/2;
									if((mid == 2 || mid== 8 || mid == 5) && mid != i1 && mid != i2 && mid != i3 && mid !=i4) continue;
									if(mid == 4 && i1 != 4 && i2!=4 && i3!=4 && i4!=4 && (i6 == 1 || i6 == 7)) continue;
									if(mid == 6 && i1 != 6 && i2!=6 && i3!=6 && i4!=6 && (i6 == 3 || i6 == 9)) continue;
								}

								if(i6 == 0){
									printf("%d\t%d%d%d%d%d\n", count + 1, i1, i2, i3, i4, i5);
									count++;
								}else{

									for(int i7=0; i7<10; i7++){
										if(i7 == i1 || i7 == i2 || i7 == i3 || i7 == i4 || i7 == i5 || i7 == i6) continue;
										if(i7!=0 && (i6+i7)%2 ==0 ){
											int mid=(i6+i7)/2;
											if((mid == 2 || mid== 8 || mid == 5) && mid != i1 && mid != i2 && mid != i3 && mid !=i4 && mid !=i5) continue;
											if(mid == 4 && i1 != 4 && i2!=4 && i3!=4 && i4!=4 && i5!=4 && (i7 == 1 || i7 == 7)) continue;
											if(mid == 6 && i1 != 6 && i2!=6 && i3!=6 && i4!=6 && i5!=6 && (i7 == 3 || i7 == 9)) continue;
										}

										if(i7 == 0){
											printf("%d\t%d%d%d%d%d%d\n", count + 1, i1, i2, i3, i4, i5, i6);
											count++;
										}else{


											for(int i8=0; i8<10; i8++){
												if(i8 == i1 || i8 == i2 || i8 == i3 || i8 == i4 || i8 == i5 || i8 == i6 || i8==i7) continue;
												if(i8!= 0 && (i7+i8)%2 ==0 ){
													int mid=(i7+i8)/2;
													if((mid == 2 || mid== 8 || mid == 5) && mid != i1 && mid != i2 && mid != i3 && mid !=i4 && mid !=i5 && mid!=i6 ) continue;
													if(mid == 4 && i1 != 4 && i2!=4 && i3!=4 && i4!=4 && i5!=4 && i6!= 4 && (i8 == 1 || i8 == 7)) continue;
													if(mid == 6 && i1 != 6 && i2!=6 && i3!=6 && i4!=6 && i5!=6 && i6!= 6 && (i8 == 3 || i8 == 9)) continue;
												}

												if(i8 == 0){
													printf("%d\t%d%d%d%d%d%d%d\n", count + 1, i1, i2, i3, i4, i5, i6, i7);
													count++;
												}else{


													for(int i9=0; i9<10; i9++){
														if(i9 == i1 || i9 == i2 || i9 == i3 || i9 == i4 || i9 == i5 || i9 == i6 || i9==i7 || i9 == i8 ) continue;
														if(i9 == 0){
															printf("%d\t%d%d%d%d%d%d%d%d\n", count + 1, i1, i2, i3, i4, i5, i6, i7, i8);
															count++;
														}else{
															printf("%d\t%d%d%d%d%d%d%d%d%d\n", count + 1, i1, i2, i3, i4, i5, i6, i7, i8, i9);
															count++;
														}

													}
												}
											}
										}
									}
								}
							}
						}
					}
				}
			}
		}
	}
	printf("\ntotal:%d\n", count);
	return 0;
}

额，看着头疼

一种是用回溯法 

#include <stdio.h>
#define bool int 
#define true 1
#define false 0

int ptn[10];
int index[10];
bool used[10];
int len;
int total;

void ini()
{
	int i;
	len = 1;
	total = 0;
	for (i = 0; i < 10; i++)
	{
		ptn[i] = 9;
		index[i] = 1;
		used[i] = false;
	}
}
bool CanUse(int len, int num)
{
	int i;
	if (len == 0)
		return false;

	for (i = 0; i < 9; i++)
		used[i] = false;

	for (i = 0; i < len - 1; i++)
		used[ptn[i] - 1] = true;

	if (used[num - 1])
		return false;

	int prenum, mid;
	prenum = ptn[len - 2];

	if (num == 5 || prenum == 5)
		return true;

	mid = (prenum + num) / 2;
	if (prenum % 2 == 1 && num % 2 == 1)
	{
		return used[mid - 1];
	}
	else if (prenum % 2 == 0 && num % 2 == 0)
	{
		if (mid == 5)
			return used[5 - 1];
	}
	return true;
}
int GetNum(int len)
{
	int i;
	for (i = index[len - 1]; i <= 9; i++)
	{
		if (CanUse(len, i))
			break;
	}
	return i;
}

bool Next()
{
	int num;
	do
	{
		num = GetNum(len);
		while (num == 10)//
		{
			index[len - 1] = 1;
			if (len > 1) index[len - 2] ++;
			len--;
			num = GetNum(len);
			if (len == 1 && num == 10 )
			return false;
		}
		
		ptn[len - 1] = num;
		index[len - 1] = num;
		len++;
		used[num] = true;
		return true;
	} while (len > 0);
	return false;
}
void print()
{
	int i;
	if (len <= 4)
		return;
	total++;
	printf("%d\t",total);
	for (i = 0; i < len - 1; i++)
	{
		printf("%d",ptn[i]);
	}
	printf("\n");
}


int main()
{
	ini();
	int t;
	while (Next())
	{
		print();
	}
	printf("%d\t", total);
	return 0;

}

效率第一种明显比第二种高，大约快了三倍 

在编程上来说，第一种速度较快，一个同事大概20分钟搞定

好几年没编程了，搞了两天才搞定，

编译后第一个程序8.5k，第二个7k

个人感觉效率快慢原因在于一是第二个函数调用多，二是第一个程序把许多判断都固化了。也不知道对不对，有高人能分析一下两个程序效率吗？

结果都一样 ，共有 389112种解锁图案。