1023. Have Fun with Numbers (20)

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

来源： <http://www.patest.cn/contests/pat-a-practise/1023>

#pragma warning(disable:4996)
#include <stdio.h>
#include <iostream>
#include <string>
using namespace std;

int main(void) {
	string s1 = "12345678901234567890",s2="12345678901234567890";
	cin >> s1;

	int n = s1.length();
	int jinwei = 0;
	for (int i = 0; i < n; i++) {
		if(s1[i]<‘5‘)
			s2[i] = 2 * (s1[i] - ‘0‘)+‘0‘;
		else {
			if (i > 0) {
				s2[i] = 2 * (s1[i] - ‘0‘) - 10+‘0‘;
				s2[i - 1] += 1;
			}
			if (i == 0) {
				s2[i] = 2 * (s1[i] - ‘0‘) - 10+‘0‘;
				jinwei = 1;
			}
		}
	
	}
	if (jinwei == 1) {
		cout << "No" << endl;
		cout << 1;
		for (int i = 0; i < n; i++) {
			cout << s2[i];
		}
		
		return 0;
	}
	int a[20] = { 0 };
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++) {
			if (s2[i] == s1[j] && a[j] == 0)
				a[j]++;
		}
	}
	int count = 0;
	for (int i = 0; i < n; i++)
		if (a[i] == 1)
			count++;
	if (count == n) {
		cout << "Yes"<<endl;
		for (int i = 0; i < n; i++) {
			cout << s2[i];
		}
	}
	else {
		cout << "No"<<endl;
		for (int i = 0; i < n; i++) {
			cout << s2[i];
		}
	}
	
	return  0;
}

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