pta--1023 Have Fun with Numbers（20 分）（hash&&字符串模拟加法运算）

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

【分析】

题意：给你一个只有1~9组成的数串，求该串×2后是不是还是有这些数量的数字组成。输出Yes或No，并输出二倍后的结果。

思路：Hash。用两个数组分别存运算之前和运算之后各个数字出现的次数。注意！最高位进位的情况！(会有两个测试点5分)。比如，9+9=18，注意首位的1 。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
using namespace std;
char s[25];
int a[10],b[10];
int main()
{
	while(~scanf("%s",s))
	{
		int flag=0,tag=1;
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		int len=strlen(s);
		for(int i=0;i<len;i++)
			a[s[i]-'0']++;
		for(int i=1;i<=9;i++)//如果有数字没用到也是不符合条件的 
			if(!a[i])tag=0;
		for(int i=len-1;i>=0;i--)
		{
			int x=2*(s[i]-'0');
			if(flag)x++,flag=0;
			if(x>9)flag=1;
			s[i]=x%10+'0';
			b[x%10]++;
		}
		for(int i=0;i<10;i++)
			if(a[i]!=b[i])
			{
				tag=0;break;
			}
		if(!tag)cout<<"No\n";
		else cout<<"Yes\n";
		len=strlen(s);
		if(flag)cout<<"1";//如果最高位有进位，因为是加法，所以最高就是加1，所以直接输出1就好了 
		for(int i=0;i<len;i++)cout<<s[i];
		cout<<endl;
	}
}