HDU 1098 Ignatius's puzzle（数论-其它）

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

111009999

 

Sample Output

22no43

 

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)=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;}