Ignatius's puzzle
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9859 Accepted Submission(s): 6898
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".
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
若a/b=x...0 称a能被b整除，b能整除a，即b|a,读作“b整除a”或“a能被b整除”。a叫做b的倍数，b叫做a的约数（或因数）。
a%b==0
摘自discuss
题目大意： 方程f(x)=5*x^13+13*x^5+k*a*x；输入任意一个数k，是否存在一个数a，对任意x都能使得f(x)能被65整除。 现假设存在这个数a ,因为对于任意x方程都成立 所以，当x=1时f(x)=18+ka 又因为f(x)能被65整出，故设n为整数 可得，f(x)=n*65; 即：18+ka=n*65; n为整数 则问题转化为， 对于给定范围的a只需要验证， 是否存在一个a使得(18+k*a)%65==0 所以容易解得 注意，这里有童鞋不理解为毛a只需到65即可 因为，当a==66时 也就相当于已经找了一个周期了，所以再找下去也找不到适当的a了
(18+k*a)%65=(18%65+k*a%65)%65;
当a=66时k*66%65==k%65(即a=1时)
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
using namespace std;
#define MA 10010
int main()
{
int n,i,k;
while(~scanf("%d",&k))
{
for(i=;i<=;i++)
{
if((+i*k)%==)
{
printf("%d\n",i);
break;
}
}
if(i>=)
printf("no\n");
} return ;
}