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<stdio.h>
#include<string.h>
int hash1[10]={0};
int hash2[10]={0};
 
int jud(char a[],int len)
{
    if(a[0]>='5'&&a[0]<='9')
    {
        return 0; //false代表不合格的进位大整数； 
    }
    if(a[0]=='0'&&len!=1)
    {
        return 2;
    }
        if(a[0]=='0'&&len==1)
    {
        return 1;
    }
    if(a[0]>='1'&&a[0]<'5')
    {
        return 1;
    }
}
int main()
{
    char a[25];
    int inta[25],i,len,temp;
    gets(a);
    len=strlen(a);
    /*将字符串数组转化为对应整数，存入变换的整数数组中*/
    for(i=0;i<len;i++)
    {
        inta[i]=a[i]-'0';
        hash1[inta[i]]++;
    }
    /*/检测输出
        for(i=0;i<len;i++)
    {
        printf("%d",inta[i]);
    }
    */
    //检测变化数组内容的数字出现次数； 
    /**/
//        printf("\n");
//        for(i=0;i<10;i++)
//    {
//        printf("%d ",hash1[i]);
//    }
    /**/
//    printf("\n");
     int count=0;
    for(i=len-1;i>0;i--)
    {
        temp=inta[i]*2+count;
        if(temp>=10)
        {
            count=1;
        }
        else{
            count=0;
        }
        inta[i]=temp%10;
    }
    inta[0]=inta[0]*2+count;
    //将变换数组*2
 
    //判断数组 
if(    jud(a,len)==1)
{
 
//对*2后的数组中数字出现的次数进行统计；     
        for(i=0;i<len;i++)
    {
        hash2[inta[i]]++;
    }
//        for(i=0;i<10;i++)
//    {
//        printf("%d ",hash2[i]);
//    }
    
    //比较两hash表是否完全相等
    for(i=0;i<10;i++)
    {
        if(hash1[i]!=hash2[i])
        {
            printf("No\n");
                for(i=0;i<len;i++)
            {
                printf("%d",inta[i]);
             }
            return 0;
        }
    }
        printf("Yes\n");
        for(i=0;i<len;i++)
    {
        printf("%d",inta[i]);
     }
}
 
else if(jud(a,len)==2)
{        
        printf("No\n");
        for(i=0;i<len;i++)
        {
            printf("%d",inta[i]);
        }
}
else
{
    printf("No\n");
            for(i=0;i<len;i++)
        {
            printf("%d",inta[i]);
        }
}
    
    
    return 0;
 }