hdu1098(基础)

Ignatius's puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8405    Accepted Submission(s): 5828

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

22no4
//hdu1098(基础&数学)//题目大意:已知f(x) = 5*x^13 + 13*x^5 + k*a*x;//现在给你给你k的值(k<10000),求a的最小整数值使得对于任意x都有f(x)的值都能整除65；//解题思路:" arbitrary integer x ,65|f(x)"主要是这句对于任意x都能整除65.由于是任意x都成立。所以可以用特殊值带入方程求解. //所以当x=1时,可得：f(1)=k*a+18.所以有18+k*a=i*65(i=1、2、3、4......).这样就可以从小到大枚举i,当找到一个//i使得a=(i*65-18)/k是一个整数时,该值就是满足上述的最小的a值. #include<cstdio>#include<cstring>int main(){int k,i,j;while(scanf("%d",&k)!=EOF){int result=-1;for(i=1;i<=10000;i++){if((i*65-18)%k==0){result=(i*65-18)/k;break;}}if(result==-1) printf("no\n");else printf("%d\n",result);}return 0;}