【PAT】A1023. Have Fun with Numbers (20)

Description:
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

//NKW 甲级练习题1034
#pragma warning(disable:4996)
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <vector>
using namespace std;
char num[21];
vector<int> vi;
int htable[10], flag = 1;
int main(){
	memset(htable, 0, sizeof(htable));
	scanf("%s", num);
	for (int i = 0; i < strlen(num); i++){
		htable[num[i] - '0']++;
		vi.push_back((num[i] - '0') * 2);
	}
	for (int i = vi.size()-1; i > 0; i--){	//这里我们不对vi[0]进位，因为如果vi[0]需要进位，那么结果一定为“No”
		if (vi[i] / 10){		//由于本题只是double，所以在一个vi[i]中不会出现三位数
			vi[i - 1] += vi[i] / 10;
			vi[i] = vi[i] % 10;
		}
		htable[vi[i]]--;
	}
	if (vi[0] / 10){			//如果最高位需要进位，即double后数字边长一位
		printf("No\n");
		for (int i = 0; i < vi.size(); i++)
			printf("%d", vi[i]);
		printf("\n");
		system("pause");
		return 0;
	}
	htable[vi[0]]--;
	for (int i = 0; i < vi.size(); i++){
		if (htable[vi[i]]){
			printf("No\n");
			flag = 0;
			break;
		}
	}
	if (flag)
		printf("Yes\n");
	for (int i = 0; i < vi.size(); i++)
		printf("%d", vi[i]);
	printf("\n");
	system("pause");
	return 0;
}