HDU_1098 Ignatius & amp; #39; s puzzle [regular question]

HDU_1098 Ignatius & #39; s puzzle [regular question]

Ignatius's puzzle
Problem DescriptionIgnatius is poor at math, he falls into ss 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 ".

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

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

Sample Input

111009999

Sample Output

22no43

F (x) = 5 * x ^ 13 + 13 * x ^ 5 + k * a * x to find the smallest a, so that for a given k, f (x) under any x can be divisible by 65.

Idea: Create a table to find the rule.

You will find that (5 * x ^ 13 + 13 * x ^ 5) % 65 every 65 x is a loop. To make the result true (k * a * x) % 65, each 65 cycles are required, and the value of a is between 0 and 64. Therefore, the following code is available:-> _->

Code:

/************************************************* 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].rsusing 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;}