【PAT】1023. Have Fun with Numbers (20)

最新推荐文章于 2024-05-20 09:00:59 发布

realxuejin

最新推荐文章于 2024-05-20 09:00:59 发布

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本文介绍了一道简单的编程题，任务是判断一个给定数字翻倍后是否仅由原数字的不同排列组成。通过使用两个数组来记录数字0到9出现的次数，实现了对输入数字的有效检查。

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题目：http://pat.zju.edu.cn/contests/pat-a-practise/1023

分析：简单题。用俩个数组标记数字0~9出现的数次就可以了。

题目描述：

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<iostream>
#include<string.h>
using namespace std;
#define max 22
char input[max];
int flag_1[11];
int flag_2[11];
int s[max];
bool isSame = true;

int main()
{
	int len,i,j,temp;
	int jinwei; //进位
	//清零
	memset(flag_1,0,sizeof(flag_1));
	memset(flag_2,0,sizeof(flag_2));

	cin>>input;
	len = strlen(input); 
	for(j=0,i=len-1; i>=0; i--,j++)
	{
		s[j] = input[i] - '0';	
		flag_1[ s[j] ] ++;
	}

	jinwei = 0; //进位	
	//乘以2
	for(i=0; i<len; i++)
	{
		temp = ( s[i]*2 + jinwei )/10;
		s[i] = (s[i]*2 + jinwei)%10;
		flag_2[ s[i] ] ++;
		jinwei = temp;
	}

    for(j=1; j<=10; j++)
	{
		if(flag_2[j] != flag_1[j])
		{
			isSame = false;
			break;
		}
	}
	if(isSame) cout<<"Yes"<<endl;
	else cout<<"No"<<endl;

	//输出
	if(jinwei > 0) { j=i; s[i] = jinwei;} //最高位有进位
	else j = --i;

	for(; j>=0; j--)
	{
		cout<<s[j];
	}
	cout<<endl;
	
	
	return 0;
}