PAT 1023

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原创最新推荐文章于 2026-04-28 06:29:34 发布 317 阅读 AI 写同款· GEO 优化 ›

PAT 1023 Have Fun with Numbers python解法 1023 Have Fun with Numbers （20 分） 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 happen... 阅读详情

#include<cstdio>
#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
#include<queue>
#include<map>
#include<set>
#include<cmath>
using namespace std;

struct bign
{
	int d[22];
	int len;
	bign()
	{
		memset(d,0,sizeof(d));
		len=0;
	}
};
string str1,str2;

bign _double(bign a)
{
	bign c;
	int carry=0;
	for(int i=0;i<str1.length();i++)
	{	
		int temp=a.d[i]+a.d[i]+carry;
		carry=temp/10;
		c.d[c.len++]=temp%10;
	}
	if(carry!=0) 
		c.d[c.len++]=carry;
	if(c.len-1>1&&c.d[c.len-1]==0)
		c.len--;
	return c;
}
void print(bign a)
{
	for(int i=0;i<a.len;i++)
	{
		printf("%d",a.d[a.len-1-i]);
	}
}
int main()
{
	bign a;
	//freopen("input.txt","r",stdin);
	cin>>str1;
	for(int i=0;i<str1.length();i++)
	{
		a.d[i]=str1[str1.length()-1-i]-'0';
	}
	bign b=_double(a);
	int hash1[10]={0};
	int hash2[10]={0};
	bool flag=false;
	if(b.len!=str1.length())
	{
		printf("No\n");
		print(b);
	}
	else
	{
		for(int i=0;i<str1.length();i++)
		{
			hash1[str1[i]-'0']++;
			hash2[b.d[i]]++;
		}
		for(int i=0;i<10;i++)
		{
			if(hash1[i]!=hash2[i])
			{
				flag=true;
			}
		}
		if(flag==true)
		{
			printf("No\n");
			print(b);
		}
		else
		{
			printf("Yes\n");
			print(b);
		}
	}
	//freopen("con","r",stdin);

	system("pause");
	return 0;
}

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