PAT1023. Have Fun with Numbers (20)

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 file 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
输入字符串统计出现的数的个数再乘以2，再统计出现数的个数比前比较
#include<stdio.h>
#include<string.h>
int main()
{
    char in[23];
    memset(in,0,sizeof(in));
    int i,num1[11],num2[11];//num1存输入每个数的个数，num2存2倍后
    while(~scanf("%s",in))
    {
        memset(num1,0,sizeof(num1));
        memset(num2,0,sizeof(num2));
        int len = strlen (in),t;
        for(i=0; i<len; i++)
            num1[in[i]-48]++;
        t=0;//表示进位
        for(i=len-1; i>=0; i--)
        {
            if((in[i]-48)*2+t>9)
            {
                in[i]=(in[i]-48)*2+t+38;
                t=1;
            }
            else
            {
                in[i]=(in[i]-48)*2+t+48;
                t=0;
            }
        }
        if(t==1)
            {
				for(i=len;i>0;i--)
					in[i]=in[i-1];
                in[0] = '1';
            }
		 
        for(i=0; i<len; i++)
            num2[in[i]-48]++;
        int kase = 1;
        for(i=0; i<10; i++)
        {
            if(num1[i]!=num2[i])
            {
                kase = 0;
                break;
            }
        }
        if(kase ==1 )
            printf("Yes\n");
        else
            printf("No\n");
        printf("%s\n",in);
        memset(in,0,sizeof(in));
    }
    return 0;
}