【PTA-A】1023 Have Fun with Numbers（大整数）

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

思路：

1.用字符串存储数字，另存数字到num数组。同时记录每个数字的次数存入a数组。

2.对num操作，从数字最后一位向前遍历，对每一位2倍后取余存入temp数组，如果≥10，前一位进一

3.遍历过程中每得出一位数字，a数组中相应次数-1

4.看a数组是否全为0，全为0说明前后个数相同

注意点： 

1.存储数字时从第1位开始，因为第0位可能会进位

2.如果长度不相同【即num[0]!=0】则为No，如果最后a数组中不为0则为No。

3.数字的第0位如果不是0要输出

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

int main() {
	string s;
	int a[15] = { 0 }, num[25] = { 0 }, temp[25] = { 0 };
	cin >> s;
	for (int i = 0; i < s.length(); i++) {
		a[s[i] - '0']++;		//记录每个数字个数
		num[i + 1] = s[i] - '0';   //存储数字，从第1位开始
	}
	for (int i = s.length(); i > 0; i--) {
		temp[i] += num[i] * 2;  //		//2倍后的数字

		if (temp[i] >= 10) {
			temp[i] %= 10;
			temp[i - 1]++;   //进位
		}
		a[temp[i]]--;    //对应数字个数减1
	}
	int flag = 1;
	for (int i = 0; i < 10; i++) {
		if (a[i] != 0) {
			cout << "No" << endl;
			flag = 0;
			break;
		}
	}
	if (flag)cout << "Yes"<<endl;
	for (int i = 0; i < s.length() + 1; i++) {
		if (i == 0 && temp[i] != 0)cout << temp[i];   //第0位判断是否为0
		if (i != 0)cout << temp[i];
	}
	return 0;
}