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)=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

Recommend

We have carefully selected several similar problems for you:  1071 1014 1052 1097 1082

题目大意：

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

解题思路：

因为 f(x+1)=5*(x+1)^13+13*(x+1)^5+k*a*x，

所以 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

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。

18+k*a=65*b

ax+by = c的方程有解的一个充要条件是：c%gcd(a, b) == 0

解题代码：“

#include <iostream>
#include <cstdio>
using namespace std;

int gcd(int a,int b){
    return b>0?gcd(b,a%b):a;
}

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;
}

HDU 1098 Ignatius's puzzle（数论-其它）