00-自测4. Have Fun with Numbers (20)

00-自测4. Have Fun with Numbers (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

8000 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

比较两个数中0~9出现的次数，数很大，可达20或21位，用字符串处理。

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

int c[10],c1[10];		//用于存放0~9出现的次数，c0是村输入的，c1是存加倍后的
int main()
{
	int i,j=0,sum=0,jinwei=0,flag=0;
	int c2[21];				//20位数的数加倍会出现21位数的情况
	string str;
	string::iterator it;
	stack<int> s;
	cin>>str;

	for(it=str.begin();it!=str.end();it++)	//字符串入栈，反序输出，便于逐个double加倍
		s.push((*it)-'0');

	while(!s.empty())
	{
		int a=s.top();
		c[a]++;
		sum=2*a+jinwei;				//字符加倍，考虑进位
		if(sum<10){
			c1[sum]++;
			c2[j++]=sum;			//逐个存放加倍后数据
			jinwei=0;
		}
		else{
			c1[sum%10]++;
			c2[j++]=sum%10;
			jinwei=1;	
		}
		s.pop();
		if(s.empty() && jinwei==1)	 //最高位有进位情况
			c2[j++]=1;
	}

	for(i=0;i<10;i++){
		if(c[i]!=c1[i])				//比较
			flag=1;
	}
	if(flag==1)
		cout<<"No"<<endl;
	else
		cout<<"Yes"<<endl;
	for(int t=j-1;t>=0;t--){		//反序输出
		cout<<c2[t];
	}
	cout<<endl;
	return 0;
}