solution.md

# 排列序列

<p>给出集合 <code>[1,2,3,...,n]</code>，其所有元素共有 <code>n!</code> 种排列。</p><p>按大小顺序列出所有排列情况，并一一标记，当 <code>n = 3</code> 时, 所有排列如下：</p><ol>	<li><code>"123"</code></li>	<li><code>"132"</code></li>	<li><code>"213"</code></li>	<li><code>"231"</code></li>	<li><code>"312"</code></li>	<li><code>"321"</code></li></ol><p>给定 <code>n</code> 和 <code>k</code>，返回第 <code>k</code> 个排列。</p><p> </p><p><strong>示例 1：</strong></p><pre><strong>输入：</strong>n = 3, k = 3<strong><br />输出：</strong>"213"</pre><p><strong>示例 2：</strong></p><pre><strong>输入：</strong>n = 4, k = 9<strong><br />输出：</strong>"2314"</pre><p><strong>示例 3：</strong></p><pre><strong>输入：</strong>n = 3, k = 1<strong><br />输出：</strong>"123"</pre><p> </p><p><strong>提示：</strong></p><ul>	<li><code>1 <= n <= 9</code></li>	<li><code>1 <= k <= n!</code></li></ul>

以下程序实现了这一功能，请你填补空白处的内容：

```cpp
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
static char *getPermutation(int n, int k)
{
	int i;
	char *result = malloc(n + 1);
	bool *used = malloc(n * sizeof(bool));
	memset(used, false, n * sizeof(bool));
	int total = 1;
	for (i = 1; i <= n; i++)
	{
		total *= i;
	}
	k = k - 1;
	for (i = 0; i < n; i++)
	{
		total /= (n - i);
		int gid = k / total;
		k %= total;
		int x = -1;
		int count = 0;
		____________________;
		used[x] = true;
		result[i] = x + 1 + '0';
	}
	result[n] = '\0';
	return result;
}
int main(int argc, char **argv)
{
	if (argc != 3)
	{
		fprintf(stderr, "Usage: ./test n, k\n");
		exit(-1);
	}
	printf("%s\n", getPermutation(atoi(argv[1]), atoi(argv[2])));
	return 0;
}
```

## template

```cpp
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
static char *getPermutation(int n, int k)
{
	int i;
	char *result = malloc(n + 1);
	bool *used = malloc(n * sizeof(bool));
	memset(used, false, n * sizeof(bool));
	int total = 1;
	for (i = 1; i <= n; i++)
	{
		total *= i;
	}
	k = k - 1;
	for (i = 0; i < n; i++)
	{
		total /= (n - i);
		int gid = k / total;
		k %= total;
		int x = -1;
		int count = 0;
		while (count <= gid)
		{
			x = (x + 1) % n;
			if (!used[x])
			{
				count++;
			}
		}
		used[x] = true;
		result[i] = x + 1 + '0';
	}
	result[n] = '\0';
	return result;
}
int main(int argc, char **argv)
{
	if (argc != 3)
	{
		fprintf(stderr, "Usage: ./test n, k\n");
		exit(-1);
	}
	printf("%s\n", getPermutation(atoi(argv[1]), atoi(argv[2])));
	return 0;
}
```

## 答案

```cpp
while (count <= gid)
{
	x = (x + 1) % n;
	if (!used[x])
	{
		count++;
	}
}
```

## 选项

### A

```cpp
while (count <= gid)
{
	x = (x + 1) % n;
	if (used[x])
	{
		count++;
	}
}
```

### B

```cpp
while (count < gid)
{
	x = (x + 1) % n;
	if (used[x])
	{
		count++;
	}
}
```

### C

```cpp
while (count <= gid)
{
	x = x % n;
	if (used[x])
	{
		count++;
	}
}
```