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$k$ 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

#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;
}