7-49 Have Fun with Numbers（20 point(s)）

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

【c语言代码】

#include
#include
int digit[10]; //静态初始化数组，默认值为0 
int main() {
  char ch[21];  //20个以内的数字 
  scanf("%s",ch);
  int len=strlen(ch), num, i, flag=0;
  for( i=len-1; i>=0; i--){
  	num=ch[i]-'0'; //char 转 int 
  	digit[num]++;
  	num=num*2+flag; //flag是进制 
  	flag=0;
  	if(num>=10){
  	    num-=10;
  	    flag=1;
	  }
	  ch[i]=(num+'0'); //int 转 char 
	  digit[num]--;
  }
  int flag1 = 0; //优化输出
  for( i = 0; i < 10; i++) {
      if(digit[i] != 0){
         flag1 = 1;
	 break;
      }
  }
  printf("%s", ( flag1 == 1) ? "No\n" : "Yes\n");
  if(flag == 1) printf("1");// 不要忘了x2后，进制可能会加一的情况（超过了20个数）
  printf("%s", ch);
  return 0;
}