大家好啊，这里是Cyber Striver的博客。今天我继续为大家带来PAT乙级的题解。

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目录

题目

题解

思路

经验总结

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题目

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题解

【2022--5--22】 WR：部分正确 16/20

#include 
int main()
{
	int n = 0,m=0,i=0,count=0;
	int a[10000] = {0};
	int b[10000] = {0};
	scanf("%d", &n);     //数组的元素个数
	scanf("%d", &m);    //向右移位数
	for (i = 0; i < n; i++)
		scanf("%d", a + i);
	for (i = 0;i<2;i++)
	{
		int mid = a[i];
		count = i;                     //走出循环的条件就是全部都已经移位了
		for (;;)
		{   
			int tem = (count - m + n * (m / n));
			if (tem < 0)
			{
				a[count] = a[count - m + n * (m / n + 1)]; b[count] = 1;
				count = count - m + n * (m / n + 1);
			}
			if (tem >= n)
			{
				a[count] = a[count - m + n * (m / n - 1)]; b[count] = 1;
				count = count - m + n * (m / n -1);
			}
			a[count] = a[count - m + n * (m / n )]; b[count] = 1;
			count = count - m + n * (m / n );
			if (b[i + m] == 1 && i + m < n)
			{
				a[i + m] = mid;
				break;
			}
			if (b[i + m - n * ((i + m) / n)] == 1 && i + m >= n)
			{
				a[i + m - n * ((i + m) / n)] = mid;
				break;
			}
		}
	}
	for (i = 0; i < n; i++)
    {printf("%d", a[i]);
    if (i != n - 1)
			printf(" ");
    }
    printf("n");
	return 0;
}

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思路

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经验总结

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本次博客到这里就结束了，如果你觉得我的文章还不错的话，请给我多多点赞评论哦！