HDUOJ-----1098 Ignatius's puzzle

Ignatius's puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5035    Accepted Submission(s): 3426

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".

 

Input

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

 

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.

 

Sample Input

11
100
9999

 

Sample Output

22
no
43

 

Author

eddy

 

 

 

 

 

 

 

 

 

 

这道题，如果能够在高中的时候，很定能够求出来的..

 

  对于           F(x)=5*x^13+13*x^5+k*a*x;

  数学归纳如下：
                              当x=1时，F(1)=5+13+k*a (此时需要找一个a使得，条件成立)。

                           假设  当x=k时，其成立；

                             当.x=k+1时，也应当成立。但实际上5*（x+1）^13+13*(x+1)^5+k*a*(x+1);

                     展开之后                 0                     13 

                                           5*（C  *x^13.........C *x^0)+13*(.....同前头.....)+k*a*x + k*a ;

                                                       13                    13                  

             最后我5*x^13+13*x^5+kax +5+13+ka  +后面部分（由于是65的倍数，就不讲啦）....由于当x=xs十，条件是成立的....所以最后只需要将（13+5+ka)%65

            等于0就行了！！而a必须小于65,不然就没有意义了！！试想a如果是65的倍数的话....那么F(x+1)处，便放到后头省掉了...

由此就不难写出代码了！！

代码如下:

 #include<iostream>
 using namespace std;
 int main( void )
 {
     int k,a;
     while(cin>>k)
     {
         a=;
       for(int i=;i<;i++)
       {
           if((+i*k)%==)
           {  a=i;
              break;
           }
       }
       if(a)cout<<a<<endl;
       else cout<<"no"<<endl;

     }
     return ;
 }