HDOJ1098

数学归纳法

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

f（x+1）= 5 * (x+1) ^ 13 + 13 * (x+1) ^ 5 + k * a * (x+1)按二项式展开

f（x+1）=5*（C(13,13)X^13+C(13,12)X^12+C(13,11)X^11+......+C(13,0)）+13*(C(5,5)X^5+C(5,4)X^4+...+C(5,0))+k * a *  (x+1)

               =f(x)+5*（C(13,12)X^12+C(13,11)X^11+......+C(13,0)）+13*(C(5,4)X^4+...+C(5,0))+k*a

               =f(x)+5*（C(13,12)X^12+C(13,11)X^11+......C(13,1)X）+13*(C(5,4)X^4+...C(5,1）X）+k*a+18

f(x）/65能整除，而多项式5*（C(13,12)X^12+C(13,11)X^11+......C(13,1)X）+13*(C(5,4)X^4+...C(5,1）X）也能整除65，所以当k*a+18能整除65时即f(x+1)能整除65

#include<iostream>

usingnamespacestd;

int main() {

int k;

while (cin>>k&&k) {

intflag =1;

for (int a= 1; a <= 65; a++) //a以65为周期

{

if (a*k % 65 == 47)

{

cout<<a<<endl;;

flag = 0;

break;

}

}

if (flag)

cout<<"no"<<endl;

}

return0;

}