PAT A 1023. Have Fun with Numbers (20)

最新推荐文章于 2022-03-03 20:54:29 发布

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最新推荐文章于 2022-03-03 20:54:29 发布

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本文介绍了一种算法，用于检查一个数字翻倍后其组成的数字是否为原数字中各数字的一种排列组合。通过统计原数字及翻倍后数字中各数字出现的次数来进行对比验证。

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题目

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

直接模拟，统计一下前后各个字符数量是否相等就可以了。

代码：

#include <iostream>
#include <string>
using namespace std;

int main()
{
	int i;
	int digit[10]={0},dig_double[10]={0};	//统计各个数字的数量
	int carry=0;	//进位标记
	char dig;
	string num,num_double;	//初始数，翻倍的数
	cin>>num;
	for(i=0;i<num.size();i++)	//统计初始数的各个数字的数量
		digit[num[i]-'0']++;
	for(i=num.size()-1;i>=0;i--)	//翻倍
	{
		dig=((num[i]-'0')*2+carry)%10+'0';
		carry=((num[i]-'0')*2+carry)/10;
		num_double.insert(num_double.begin(),dig);
	}
	if(carry==1)	//如果最后有进位
		num_double.insert(num_double.begin(),'1');

	for(i=0;i<num_double.size();i++)	//统计翻倍后的各个数字数
		dig_double[num_double[i]-'0']++;

	for(i=0;i<10;i++)	//比较各个数字数是否相同
		if(digit[i]!=dig_double[i])
		{
			cout<<"No"<<endl;
			cout<<num_double;
			return 0;
		}
	cout<<"Yes"<<endl;
	cout<<num_double;

	return 0;
}