标签：acm c++ 杭电 算法 编程
Ignatius‘s puzzle
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7068 Accepted Submission(s): 4883
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
Sample Output
Author
eddy
题目的关键是函数式f(x)=5*x^13+13*x^5+k*a*x;
用数学归纳法证明：x取任何值都需要能被65整除..
所以我们只需找到f(1)成立的a,并在假设f(x)成立的基础上,
证明f(x+1)也成立即可。此时经过整理 只要（18+k*a）能够被65整除 ，求最小a的问题。
#include<iostream>
using namespace std;
int main()
{
int k,a;
while(cin>>k)
{
int i;
for( i=0;i<66;i++)
if((18+k*i)%65==0)
{
cout<<i<<endl;
break;
}
if(i==66)
cout<<"no"<<endl;
}
return 0;
}
杭电 HDU 1098 Ignatius's puzzle
标签：acm c++ 杭电 算法 编程
原文地址：http://blog.csdn.net/lsgqjh/article/details/44685981
