[C++知识库]【中国大学MOOC-陈越、何钦铭-数据结构-起步能力自测题】自测4 Have Fun with Numbers

Question：
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
结尾无空行

Sample Output:

Yes
2469135798
结尾无空行

代码

#define _CRT_SECURE_NO_WARNINGS 1 
#include<stdio.h>
#include <string.h>
int judge(int arr1[],int arr2[],int x)
{
	for (int i = 0; i < x; i++)
	{
		if (arr1[i] != 0 && arr2[0] != 0)
		{
			return 0;
		}
	}
	return 1;
}
int main()
{
	char temp1[22];//方便存入

	scanf("%s", temp1);
	int sz = strlen(temp1);//计算位数
	int a[50] = { 0 };
	int b[51] = { 0 };
	int dit[10] = { 0,0,0,0,0,0,0,0,0,0 };

	for (int i = 0; i < sz; i++)
	{
		a[i] = temp1[i] - '0';	//temp中的数字存入整型数组
		dit[a[i]]++;//记录0-9数字分别的个数
	}
	int carry_bit = 0;
	for (int i = sz - 1; i >= 0; i--)
	{
		b[i+1] =carry_bit;
		int amass= a[i] * 2 ;
		if (amass >=10)
		{
			b[i+1] = amass % 10+b[i+1];
			carry_bit = (amass -b[i+1]+ carry_bit)/10;
			if (b[i+1] >= 10)
			{
				carry_bit++;
				b[i+1] = b[i+1] - 10;
			}

		}
		else
		{
			b[i+1] = amass+b[i+1];
			carry_bit = 0;
		}

	}
	b[0] = carry_bit;
	for (int i = 0; i < sz; i++)
	{
		dit[b[i+1]]--;
	}

	if (judge(dit,b,sz)== 0)
	{
		printf("No\n");
		for (int i = 0; i < sz + 1; i++)
		{
			printf("%d", b[i]);
		}
	}
	else
	{
		printf("Yes\n");
		for (int i = 1; i < sz + 1; i++)
		{
			printf("%d", b[i]);
		}
		
	}

	return 0;
}