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): 5765    Accepted Submission(s): 3968

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
   
   
   
   

数论题，整整一节课都是它的了，还参看了很多orz的博客，不过也还是学到了点东西.

考虑特殊值，当X=1时，就可以得到那什么东西：

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：假设存在这个数a ,因为对于任意x方程都成立，所以，18+ka=n*65;若该方程有整数解则说明假设成立。

对于方程有整数解：a*x+b*y=m;如a，b的最大公约数为1，则有整数解

或

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

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int gcd(int a, int b)
{
    return b? gcd(b, a%b):a;
}
int main()
{
    int m;
    while(~scanf("%d", &m))
    {
        if(gcd(65, m)==1)
        {
            for(int i = 1;; i++)
            {
                if((i*65-18)%m == 0)
                {
                    printf("%d\n", (i*65-18)/m);
                    break;
                }
            }
        }
        else printf("no\n");
    }
    return 0;
}