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最新推荐文章于 2024-05-20 09:00:59 发布

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最新推荐文章于 2024-05-20 09:00:59 发布

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本文介绍了一个编程问题：检查一个给定的数字，在将其翻倍之后，所得的新数字是否仅由原数字的位数重新排列组成。文章提供了一段C语言代码实现，该程序能够接收一个不超过20位的正整数作为输入，并判断翻倍后的数字是否满足题目要求。

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1023. Have Fun with Numbers (20)

时间限制

400 ms

内存限制

32000 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 <stdio.h>
#include <string.h>

char str[25];
int num[10], ans[25], num1[10];

int main()
{
	int i = 0, tmp, j;

	while(scanf("%s", str) != EOF)
	{
		memset(num, 0, sizeof(num));
		memset(ans, 0, sizeof(ans));
		memset(num1, 0, sizeof(num1));

		int len = strlen(str);

		for(i = 0; i < len; i++)
				num[str[i] - '0']++;

		for(i = len - 1, j = 0; i >= 0; i--, j++)
		{
				int tmp = str[i] - '0';
				tmp = 2 * tmp;
				ans[j] = tmp;
		}

		for(i = 0; i < j; i++)
		{
			if(ans[i] >= 10)
			{
				ans[i] = ans[i] % 10;
				ans[i+1]++;
			}
		}

		if(ans[i] == 0)
			i--;

		for(j = 0; j <= i; j++)
		{
			num[ans[j]]--;
		}

		for(i = 0; i < 10; i++)
			if(num[i] != 0)
				break;

		if(i < 10)
			printf("No\n");
		else
		{
			printf("Yes\n");
		}
		for(i = j - 1; i >= 0; i--)
			printf("%d", ans[i]);
		printf("\n");
	}
	return 0;
}