HDU1098 Ignatius's puzzle(数学归纳法求解）（数论）

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

题目大意：输入k,然后求对于任意的x，使f（x）能被65整除时，输出此时最小的a。

想法如下：
数学归纳法证明：

f(x)=5*x^13+13*x^5+k*a*x

第一步：f(0)=0;成立：

第二步：假设f(x)能被65整除，则有5*x^13+13*x^5+k*a*x能被65整出；

第三步：则f(x+1)=5*(x+1)^13+13*(x+1)^5+k*a*(x+1);

根据二项式定理分析题目：

f(x+1)=5*(c(13,0)+c(13,1)x+c(13,2)*x^2+…….+c(13,13)*x^13)+13(c(5,0)+c(5,1)x+……+c(5,5)*x^5)+k*a(x+1)

=f(x)+5*(c(13,0)+c(13,1)x+c(13,2)*x^2+…….+c(13,12)*x^12)+13(c(5,0)+c(5,1)*x+……+c(5,4)*x^4)+k*a;

=f(x)+5+5*c(13,1)*x+5*c(13,2)*x^2+……+5*c(13,12)*x^12+13+13*c(5,1)*x+…..+k*a;
此时会发现5(n+1)^13+13(x+1)^5一定能被%65==0.

用f(x+1)-f(x),会发现，只有18+ka不能确定是否能%65==0；又因（18+ka)%65=(18%65+(k%65)*(a%65))%65,由此可以确定0

#include 
int main()
{
    int k,i;
    while(~scanf("%d",&k))
    {

        for(i=1;i<=65;i++)
        {
            if((18+k*i)%65==0)
            {
                printf("%d\n",i);
                break;
            }
        }
        if(i==66)printf("no\n");
    }

    return 0;

}