数学--数论--HDU 1098 Ignatius's puzzle （费马小定理+打表）

风骨散人Chiam

发布于 2020-11-06 00:02:32

5400

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)=5x13+13*x5+ka*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

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题目大意：

给定一个k，找到最小的a 使得 f(x)=5x13+13*x5+ka*x ，f(x)%65永远等于0

打表的话就很明显的看导规律

也可以用费马小定理证明

#include <iostream>
#include <cstdio>
using namespace std;
int gcd(int a, int b)
{
    if (a < b)
        return gcd(b, a);
    if (b == 0)
        return a;
    if ((a & 1) == 0 && (b & 1) == 0)
        return 2 * gcd(a >> 1, b >> 1); //a and b are even
    if ((a & 1) == 0)
        return gcd(a >> 1, b); // only a is  even
    if ((b & 1) == 0)
        return gcd(a, b >> 1);              // only b is  even
    return gcd((a + b) >> 1, (a - b) >> 1); // a and b are odd
}
int main()
{
    int k;
    while (scanf("%d", &k) != EOF)
    {
        if (18 % gcd(k, 65) == 0)
        {
            for (int a = 0;; a++)
            {
                if ((18 + k * a) % 65 == 0)
                {
                    printf("%d\n", a);
                    break;
                }
            }
        }
        else
            printf("no\n");
    }
    return 0;
}

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原始发表：2020/01/26 ，如有侵权请联系 cloudcommunity@tencent.com 删除

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