PAT 1023 Have Fun with Numbers (大数相乘)

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

大数相乘

代码

#include <stdio.h>
#include <string>
#include <stdlib.h>
#include <iostream>
#include <vector>
#include <string.h>
#include <algorithm>
#include <cmath>
#include <map>
#include <queue>
#include <stack>
#include <functional>
#include <limits.h> 
using namespace std;


int main() {
	string a;
	cin >> a;
	reverse(a.begin(), a.end());
	int num[100];	memset(num, 0, sizeof(num));
	string ans = "";
	int carry = 0;
	int flag = true;
	for(int i = 0; i < a.length(); i++){
		num[a[i] - '0']++;
		ans.push_back((2 * (a[i] - '0') + carry) % 10  + '0');
		carry = (2 * (a[i] - '0') + carry) / 10;
	}
	if(carry){
		ans.push_back(carry + '0');
	}
	for(int i = 0; i < ans.length(); i++){
		num[ans[i] - '0']--;
	}
	for(int i = 0; i < 15; i++){
		if(num[i] != 0){
			flag = false;
			break;
		}
	}
	reverse(ans.begin(), ans.end());
	if(flag)	cout << "Yes" << endl;
	else cout << "No" << endl;
	cout << ans << endl;
	return 0; 
}