杭电acm 1098题

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

 本题大意是要求对任意输入的K，找到一个最小的a，使得表达式都能整除65.

自己的做法是在一定范围内去找，1-10000之间找到当x=1时能够整除65的，然后在这基础上让x在1-10000之间去循环整除，如果能满足所有条件那么就找到了这个最小的a。否则就输出no。程序如下，已经AC....

/*******************************************************
杭电acm  1098 已经AC
*******************************************************/

#include <iostream>
#include <cmath>
#define fx 5*pow(x,13)+13*pow(x,5)+k*a*x
using namespace std;

int main(void)
{
    int x=1;
    int k;
    int a=1;
    //long int fx;
    int mark=1;
    int flag=0;
    while(scanf("%d",&k)!=EOF)
    {
        mark=1;
        flag=0;
        //fx=5*pow(x,13)+13*pow(x,5)+k*a*x;
        for(a=1;a<10000;a++)
        {
            if((18+k*a)%65==0)
            {
                for(x=1;x<1000;x++)
                {
                    if(fx%65==0)
                        continue;
                    else {break;mark=0;}
                }

            }
            if(((18+k*a)%65==0)&&mark==1)
            {flag=1;break;}
        }
        if(mark&&flag)
        {
            cout<<a<<endl;
        }
        else cout<<"no"<<endl;

    }
    return 0;
}