hdu 1098 Ignatius's puzzle （数论）

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

 转载自：http://blog.sina.com.cn/s/blog_5fe933490100ey6r.html

题目的关键是f(x)=5*x^13+13*x^5+k*a*x;
由于x取任何值都需要能被65整除.那么用数学归纳法.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.
那么把f(x+1)展开,得到5*( ( 13 0 )x^13 + (13 1 ) x^12 ...... .....+(13 13) x^0)+13*( ( 5 0 )x^5+(5 1 )x^4......其实就是二项式展开,这里就省略了 ......+ ( 5 5 )x^0 )+k*a*x+k*a;——————这里的( n m)表示组合数,相信学过2项式定理的朋友都能看明白.

然后提取出5*x^13+13*x^5+k*a*x
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.

 

代码如下：

#include<bits/stdc++.h>
using namespace std;
#define ll long long

int main()
{
    int k,a,res;
    while(cin >> k)
    {
        bool flag = 0;
        for(int a = 0;a < 66;a++)
        {
            res = 18 + k * a;
            if(res % 65 == 0)
            {
                cout << a << endl;
                flag = 1;
                break;
            }
        }
        if(flag == 0)
        cout << "no" << endl;
    }
    return 0;
}