PAT甲级1023

最新推荐文章于 2024-05-20 09:00:59 发布

strivinging

最新推荐文章于 2024-05-20 09:00:59 发布

阅读量847

点赞数 1

CC 4.0 BY-SA版权

分类专栏： PAT 文章标签： C++ C++程序设计程序设计

本文链接：https://blog.youkuaiyun.com/qq_22194315/article/details/54684546

PAT 专栏收录该内容

200 篇文章

订阅专栏

本文介绍了一种算法，用于检查给定数字翻倍后是否仅由原始数字的排列组成。通过对输入数字进行位运算和哈希表记录，该算法能够高效判断并输出结果。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1023. Have Fun with Numbers (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9, with no duplication. Double it we will obtain 246913578, which happens to be another 9-digit number consisting exactly the numbers from 1 to 9, only in a different permutation. Check to see the result if we double it again!

Now you are suppose to check if there are more numbers with this property. That is, double a given number with k digits, you are to tell if the resulting number consists of only a permutation of the digits in the original number.

Input Specification:

Each input file contains one test case. Each case contains one positive integer with no more than 20 digits.

Output Specification:

For each test case, first print in a line "Yes" if doubling the input number gives a number that consists of only a permutation of the digits in the original number, or "No" if not. Then in the next line, print the doubled number.

Sample Input:

1234567899

Sample Output:

Yes
2469135798

#include<cstdio>
#include<vector>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
	char s[22];
	scanf("%s", s);
	int hash[10] = { 0 };
	int len = strlen(s);
	for (int i = 0; i < len; i++)
	{
		hash[s[i] - '0']++;
	}
	int t, carry = 0,dt;
	bool flag = false;
	for (int i = len - 1; i >= 0; i--)
	{
		t = s[i] - '0';
		dt = t << 1;
		if (dt >= 10)
		{
			s[i] = '0' + dt - 10+carry;
			carry = 1;
		}
		else
		{
			s[i] = '0' + dt+carry;
			carry = 0;
		}
		if (i == 0 && carry)
			flag = true;
	}
	for (int i = 0; i < len; i++)
	{
		hash[s[i] - '0']--;
	}
	
	if (flag)
	{
		printf("No\n");
		printf("1");//这个地方得注意，由于不管是No,还是Yes都要输出翻倍后的数，即使数字长度不同
		printf("%s", s);
		return 0;
	}
	else
	{
		for (int i = 0; i < len; i++)
		{
			if (hash[s[i] - '0'] != 0)
			{
				printf("No\n");
				printf("%s", s);
				return 0;
			}
		}
	}
	printf("Yes\n");
	printf("%s", s);
	return 0;
}
/*一定要先弄清楚题意，不要有个大概的思路就开始写。题目完全弄懂，把可能出现的坑点想清楚，算法思想一定要清晰然后在开始动手写，模糊的话就不要开始写
因为最后会漏洞百出。*/