数据结构起步能力自测4-have fun with numbers

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
kkk
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

#include <iostream>
#include <cmath>
#include <string.h>
using namespace std;
#define max 22

char input[max];
int flag_1[11];
int flag_2[11];
int s[max];
bool isSame=true;

int main()
{
int len,i,j,temp;
int jinwei;//进位
//清零
memset(flag_1,0,sizeof(flag_1));
memset(flag_2,0,sizeof(flag_2));

cin>>input;
len=strlen(input);
for (j=0,i=len-1;i>=0;i--,j++)//i--倒叙输入，低位数放前面，在编写加法的时候逻辑比较简单
{
s[j]=input[i]-'0';//将char数字转换为int数字
flag_1[s[j]]++;//统计输入数据中数字出现频率
}
jinwei=0;
//编写乘法
for(i=0;i<len;i++)
{
temp=(s[i]*2+jinwei)/10;
s[i]=(s[i]*2+jinwei)%10;
flag_2[s[i]]++;//统计输出数据中数字出现的频率
jinwei=temp;
}

for (j=1;j<=10;j++)
{
if(flag_1[j]!=flag_2[j])
{
isSame=false;
break;
}
}
if (isSame)
cout<<"Yes"<<endl;
else
cout<<"No"<<endl;

//输出数字
if (jinwei>0)//最高位有进位
{
j=i;
s[i]=jinwei;//最高位为进位
}
else
{
j=--i;
}

for(;j>=0;j--)//同样倒叙输出
{
cout<<s[j];
}
cout<<endl;
return 0;
}