Ignatius's puzzle

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4935????Accepted Submission(s): 3359

Problem Description

Ignatius is poor at math,he falls across a puzzle problem,so he has no choice but to appeal to Eddy. this problem describes that:f(x)=5*x^13+13*x^5+k*a*x,input a nonegative integer k(k<10000),to find the minimal nonegative integer a,make the arbitrary integer x ,65|f(x)if
no exists that a,then print "no".

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Input

The input contains several test cases. Each test case consists of a nonegative integer k, More details in the Sample Input.

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Output

The output contains a string "no",if you can't find a,or you should output a line contains the a.More details in the Sample Output.

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Sample Input

11 100 9999

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Sample Output

22 no 43

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Author

eddy

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若對(duì)任意x成立,則當(dāng)x=1時(shí)必然成立.若當(dāng)x=1時(shí)成立,則對(duì)任意正整數(shù)x都成立.

觀察題目中給出的式子可以看出每一項(xiàng)都有公因子x,所以f(x)必然是f(1)的x倍,若f(1)是65的倍數(shù),則f(x)必然也是65的倍數(shù).

      #include<stdio.h>


      
        int
      
      
         main()

{

    
      
      
        int
      
      
         k;

    
      
      
        while
      
       (scanf(
      
        "
      
      
        %d
      
      
        "
      
      ,&k)!=
      
        EOF)

    {

        
      
      
        if
      
       (k==
      
        1
      
       || (k%
      
        5
      
      !=
      
        0
      
       && k%
      
        13
      
      !=
      
        0
      
       && k%
      
        65
      
      !=
      
        0
      
      
        ))

        {

            
      
      
        for
      
       (
      
        int
      
       i=
      
        1
      
      ;;i++
      
        )

            
      
      
        if
      
       ((
      
        18
      
      +k*i)%
      
        65
      
      ==
      
        0
      
      
        )

            {

                printf(
      
      
        "
      
      
        %d\n
      
      
        "
      
      
        ,i);

                
      
      
        break
      
      
        ;

            }

        }

        
      
      
        else
      
       printf(
      
        "
      
      
        no\n
      
      
        "
      
      
        );

    }

    
      
      
        return
      
      
        0
      
      
        ;

}
      
    

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Ignatius's puzzle