hdu 1098 Ignatius's puzzle

Ignatius's puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6148    Accepted Submission(s): 4231

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

 

假设存在这个数a ,因为对于任意x方程都成立，所以，当x=1时f(x)=18+ka;有因为f(x)能被65整出，这可得出f(x)=n*65;

即：18+ka=n*65;若该方程有整数解则说明假设成立。

二元一次方程整数解存在的条件：在整系数方程ax+by=c中，  若a,b的最大公约数能整除c,则方程有整数解。即 如果（a,b）|c 则方程ax+by=c有整数解  显然a,b互质时一定有整数解。

#include <iostream>#include<stdio.h>using namespace std;int gcd(int a,int b){    if(b==0)    {        return a;    }    else    {        return gcd(b, a%b);    }}int main(){    int k;    while(cin>>k)    {        if(18%gcd(65,k)==0)        {            for (int i=1;; i++)            {                if((i*65-18)%k==0)                {                    cout<<(i*65-18)/k<<endl;                    break;                }            }        }        else        {            cout<<"no"<<endl;        }    }    return 0;}