C-The story of the Fox and the Rabbit

Title Description

Around the top of the mountain there are 10 holes, a fox and a rabbit each live a hole. The fox always wants to eat rabbits. One day the rabbit said to the Fox: "You want to eat I have a condition, first the hole from the 1~10 number, you start from hole 10th, first to 1th hole looking for me, the second time 1 holes to find me, the third time 2 holes to find me, and so on, and so on. If you can find me, you can have a hearty meal. But I can't stop until I find me. "The fox is full of promise to start looking for, it from morning to night to find n times hole, tired faint past also did not find rabbits." Excuse me, which hole did the fox faint in? What holes can the child hide in?

Input

Enter a positive integer N (1<=n<=1000)

Output

First line: Output the number of holes the rabbit can hide in
Second line: The number of the hole where the fox fainted

Sample input
3
Sample output
2 4 5 7 8 9 10
6

1 /************2 variable definition:3 N: Find the N-th hole altogether4 K: K-Find Hole (1<= k <=n)5 6 find_rabbit: Returns the numbered of the hole found in the K-th. 7 8 **********/9#include <stdio.h>Ten intFind_rabbit (intk) { One     if(k==1){ A         return 1;/*first time in hole 1th,*/ -}Else{ -         if(Find_rabbit (K-1) + k)%Ten==0){ the             return Ten; -}Else{ -             return(Find_rabbit (K-1) + k)%Ten; -         } +     } - } + voidMain () { A     inti,holes[Ten]; at     intn= -; -      for(i=0; i<Ten; i++){ -Holes[i] =1;/*initialized to 1, 1 for the Fox has not been to the hole*/ -     } -      -      for(i=1; i<=n; i++){ inHoles[find_rabbit (i)-1] =0;/*eg: the first time in hole 1th, that is holes[0] is set to 0, use index to reduce 1, 0 for the fox to the hole*/ -     } to  +      for(i=0; i<Ten; i++){ -         if(Holes[i] = =1){ theprintf"%d", i+1);/*because I is starting from 0, so add 1*/ *         } $     }Panax Notoginsengprintf"\n%d \ n", Find_rabbit (n)); -  the      +}

C-The story of the Fox and the Rabbit