Have Fun with Numbers

最新推荐文章于 2020-06-04 15:44:32 发布

jgzquanquan

最新推荐文章于 2020-06-04 15:44:32 发布

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文章标签： PAT c语言算法编程

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本文介绍了一个有趣的数学问题：寻找那些在翻倍后其数字仅是原数字排列组合的整数。通过提供一个完整的C语言程序，展示了如何验证一个给定的数字是否具有这种特性，并附带了详细的代码实现。

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

提交代码

#include<stdio.h>
#include<string.h>
int r;
struct bign{
	int d[1000];
	int len;
	bign()
	{
		memset(d,0,sizeof(d));
		len=0;
	}
};
bign change(char str[])
{
	bign a;
	a.len=strlen(str);
	for(int i=0;i<a.len;i++)
	{
		a.d[i]=str[a.len-1-i]-'0';
	}
	return a;
}
bign multi(bign a,int b)
{
	bign c;
	int carry=0;
	for(int i=0;i<a.len;i++)
	{
		int temp=a.d[i]*b+carry;
		c.d[c.len++]=temp%10;
		carry=temp/10;
	}
	while(carry!=0)
	{
		c.d[c.len++]=carry%10;
		carry/=10;
	}
	return c;
}
void prin(bign a)
{
	for(int i=a.len-1;i>=0;i--)
	{
		printf("%d",a.d[i]);
	}
}
int judge(bign a,bign b)
{
	if(a.len!=b.len) return 0;
	else
	{
		int counta[10]={0},countb[10]={0};
		for(int i=0;i<a.len;i++)
		{
			counta[a.d[i]]++;
			countb[b.d[i]]++;
		}
		for(int i=0;i<10;i++)
		{
			if(counta[i]!=countb[i]) return 0;
		}
		return 1;
	}
}
int main()
{
	char str[30];
	scanf("%s",str);
	bign a=change(str);
	bign b=multi(a,2);
	if(judge(a,b)==1) printf("Yes\n");
	else printf("No\n");
	prin(b);
	return 0;
}