hdu 1098 Ignatius's puzzle-优快云博客

本文介绍了一种通过编程解决特定数学问题的方法。该问题要求找到最小的非负整数a，使得对于任意整数x，函数f(x)=5*x^13+13*x^5+k*a*x的值能被65整除。若不存在这样的a，则输出no。通过数学归纳法证明了求解过程，并提供了一个简单的C++程序实现。

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

解答：

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。
那么把f(x+1)展开，则f(x+1 ) = f (x) + 5*( (13 1 ) x^12 ...... .....+(13 13) x^0 )+ 13*( (5 1 )x^4+...........+ ( 5 5 )x^0 )+k*a;

很容易证明，除了5*(13 13) x^0 、13*( 5 5 )x^0 和k*a三项以外,其余各项都能被65整除.
那么也只要求出18+k*a能被65整除就可以了.
而f(1)也正好等于18+k*a

所以,只要找到a,使得18+k*a能被65整除,也就解决了这个题目.

代码：

#include<iostream>
using namespace std;
int main()
{
	int k,i;
	while(scanf("%d",&k)!=EOF)
	{
		for(i=1;i<=10000;i++)
		{
			if((18+k*i)%65==0){printf("%d\n",i);break;}
		}
		if(i>10000)printf("no\n");
	}
	return 0;
}