PAT (Advanced Level) Practice 1023 Have Fun with Numbers （20 分）

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 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 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

Code

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

const int MAXN = 25;
int a[MAXN], b[MAXN];

void big_add(string s0, string s1)
{
	memset(a, 0, sizeof(a));
	memset(b, 0, sizeof(b));
	for (int i = 0; i < s0.length(); i++)
		a[i] = (s0[s0.length() - i - 1] - '0');
	for (int i = 0; i < s1.length(); i++)
		b[i] = (s1[s1.length() - i - 1] - '0');
	int plus = 0;
	for (int i = 0; i < MAXN; i++)
	{
		int sum = a[i] + b[i] + plus;
		plus = sum / 10;
		a[i] = sum % 10;
	}
}

int main()
{
	string s0;
	cin >> s0;
	int len;
	big_add(s0, s0);
	for (int i = MAXN - 1; i >= 0; i--)
		if (a[i]) { len = i + 1; break; }
	int acopy[MAXN];
	int acopyy[MAXN];
	for (int i = 0; i < len; i++)
	{
		acopy[i] = a[len - 1 - i];
		acopyy[i] = a[len - 1 - i];
	}
	sort(s0.begin(), s0.end());
	sort(acopyy, acopyy + len);
	bool but = true;
	for (int i=0 ; i<len; i++)
		if (s0[i] - '0' - acopyy[i]) { but = false; break; }
	cout << (but ? "Yes\n" : "No\n");
	for (int i = 0; i < len; i++)
		cout << acopy[i];
	cout << '\n';
	return 0;
}

思路

本来用Long Long但是过不了，所以大数加。