hdu1098Ignatius's puzzle

Ignatius's puzzle

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

111009999

 

Sample Output

22no43

 

/*数学归纳法 题意：方程f(x)=5*x^13+13*x^5+k*a*x       输入任意一个数k（k>=0&&k<10000)是否存在一个数a，对任意x都能使得f(x)被65整除。 做法：这个题目的话肯定是有巧妙方法，思维题目      f(x)能被65整除 说明f(x)=65*n（n为整数）  即 f(x)=5*x^13+13*x^5+k*a*x=65*n   根据数学归纳法 令x=1 则 18+ka = 65*n   只需找到一个最小的a满足 (18+ka)%65==0 即可。      因为 k 、a 都是非负的整数 所以 a>=65的时候还没找到就输出no */#include <cstdio>#include <cstring>#include <cmath>#include <iostream>#include <algorithm>#include <string>#include <cstdlib>using namespace std;int main(void){int k;while(scanf("%d",&k)!=EOF){int flag=0;for(int i=0;i<65;i++){if((18+k*i)%65==0){flag=1;printf("%d\n",i);break;}}if(flag==0) printf("no\n");} return 0;}