其实和打表差不多，打表的计算在低下自己算
这题没别的方法
因为数据很多，虽然只有13种情况，但是它不停的重复
> int main(void)
> {
> 	int k;
> 	for(;;)
> 	{
> 		scanf("%d", &k);
> 		if(k == 0)
> 		{
> 			break;
> 		}
> 		__asm
> 		{
> 			mov edi, dword ptr [k]
> 			mov ebx, edi
> 			shl edi, 1
> 		_OLoop:
> 			mov ecx, edi
> 			mov eax, ebx
> 			xor edx, edx
> 			dec eax
> 			div ecx
> 			mov eax, edx
> 		_ILoop:
> 			mov edx, edi
> 			shr edx, 1
> 			cmp eax, edx
> 			jge _2ndBranch
> 		_1stBranch:
> 			cmp ecx, edx
> 			je _Found
> 			inc ebx
> 			jmp _OLoop
> 		_2ndBranch:
> 			dec ecx
> 			dec eax
> 		_AfterBranches:
> 			add eax, ebx
> 			xor edx, edx
> 			div ecx
> 			mov eax, edx
> 			jmp _ILoop
> 		_Found:
> 			mov dword ptr [k], ebx
> 		}
> 		printf("%d\n", k);
> 	}
> }
> 
> 汇编那段相当于
> (ecx = n, ebx = m, eax = i, edi = k*2)
> int m, f = 0;
> for(m = k; f == 0; m++)
> {
> 	int n = k;
> 	int i;
> 	for(i = (m - 1) % n; ; i = (i + m) % n)
> 	{
> 		if(i < k >> 1)
> 		{
> 			if(n == k >> 1)
> 			{
> 				f = 1;
> 			}
> 			break;
> 		}
> 		else
> 		{
> 			n--;
> 			i--;
> 		}
> 	}
> }