HDU 1098 Ignatius's puzzle 【数学(特殊值总结) Or 规律（瞎搞）】

[align=left]Problem Description[/align]
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".

 

[align=left]Input[/align]
The input contains several test cases. Each test case consists of a nonegative integer k, More details in the Sample Input.

 

[align=left]Output[/align]
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.

 

[align=left]Sample Input[/align]

11
100
9999

 

[align=left]Sample Output[/align]

22
no
43

题意：f(x)=5*x^13+13*x^5+kax，输入一个数k，判断它是否满足存在a，使x取任意值满足f(x)为65的倍数，存在则输出最小的a，不存在则输出no

思路：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;因为n为整数，若要方程成立则问题转化为，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，这里有童鞋不理解怎么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）

AC代码：

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

int main()
{
int n, ans;
while(~scanf("%d",&n))
{
ans = 0;
for(int i = 1; i <= 65; ++i)
{
if((18 + n * i) % 65 == 0)
{
ans = i;
break;
}
}
if(ans) cout << ans << endl;
else cout << "no" << endl;
}
return 0;
}