Ignatius's puzzle

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

题意：f(x)=5*x^13+13*x^5+k*a*x,找出最小的a，使得对于给定的k，任意x下f（x）能被65整除。

思路：打表找规律。

你会发现，（5*x^13+13*x^5）%65每65个x成循环。要使结果成立（k*a*x）%65，也得每65个循环，且a的值在0-64之间，所以就有如下代码-> _ ->

代码：

/************************************************
* Author: Ac_sorry
* File:
* Create Date:
* Motto: One heart One life
* CSDN: http://blog.csdn.net/code_or_code
*************************************************/

#include
  
   
#include
   
     #include
    
      #include
     
       #include
      
        #include
       
         #include
        
          #include
         
           #include
           #include
           
             #include
            
              #include
             
               #define INF 0x3f3f3f3f #define MOD 1000000007 #define seed_ 131 #define eps 1e-8 #define mem(a,b) memset(a,b,sizeof a) #define w(i) T[i].w #define ls(i) T[i].ls #define rs(i) T[i].rs using namespace std; typedef long long LL; const int N=100010; int a[100]; int main() { //int T;scanf("%d",&T); //int AC=0; for(int x=1;x<=65;x++) a[x]=(5*x%65*x*x%65*x*x%65*x*x%65*x*x%65*x*x%65*x*x%65+13*x%65*x*x%65*x*x%65)%65; int k; while(scanf("%d",&k)==1) { int ok,ans; for(int aa=0;aa<65;aa++) { ok=1; for(int i=1;i<=65;i++) { if((a[i]+k*i%65*aa%65)%65!=0) { ok=0; break; } } if(ok) { ans=aa; break; } } if(ok) printf("%d\n",ans); else printf("no\n"); } return 0; }

HDU_1098 Ignatius's puzzle[规律题]

Ignatius's puzzle