Ignatius‘s puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

题目的关键是函数式f(x)=5*x^13+13*x^5+k*a*x;  
用数学归纳法证明：x取任何值都需要能被65整除..

所以我们只需找到f(1)成立的a,并在假设f(x)成立的基础上,

证明f(x+1)也成立即可。此时经过整理 只要（18+k*a）能够被65整除 ，求最小a的问题。

#include<iostream>
using namespace std;
int main()
{
	int k,a;
	while(cin>>k)
	{
		int i;
		for( i=0;i<66;i++)
 		  if((18+k*i)%65==0)
		  {
			  cout<<i<<endl;
			break;
		  }
		if(i==66)
			cout<<"no"<<endl;
	}
	return 0;
}

杭电 HDU 1098 Ignatius's puzzle