7-2 Have Fun with Numbers (20分)

7-2 Have Fun with Numbers (20分)

 

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



代码的讲解，本题主要利用了hash和大数相乘。想明白这俩个就很容易了


#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define maxsize 30 
int main()
{
 char a[maxsize];
 int b[maxsize];
 int c[10];
 int d[10];
 int dig=0;
 memset(b,0,maxsize*sizeof(int));
 memset(c,0,10*sizeof(int));
 memset(d,0,10*sizeof(int));
 scanf("%s",a);
 if(a[0]=='0')
 {
  printf("Yes\n");
  printf("0");
  return 1;
 }
 int i;
 int str=strlen(a);
 for(i=str-1;i>=0;i--)
 {
  c[a[i]-'0']++;
  b[str-i-1]=(a[i]-'0')*2+dig;
  if(b[str-i-1]>9)
  {
   dig=b[str-i-1]/10;
   b[str-i-1]=b[str-i-1]%10;
  }
  else
  {
   dig=0;
  }
 }
 if(dig)
 {
  b[str]=dig;
 }
 for(i=maxsize-1;i>=0;i--)
 {
  if(b[i]!=0)
  {
   break;
  }
 }
 int j;
 for(j=i;j>=0;j--)
 {
  d[b[j]]++;
 }
 for(j=0;j<10;j++)
 {
  if(c[j]!=d[j])
  break;
 }
 if(j==10)
 {
  printf("Yes\n");
  
 }
 else
 printf("No\n");
 for(j=i;j>=0;j--)
 printf("%d",b[j]);
 printf("\n");
 
 
 
 
 
 
 return 0;
 }