杭电 HDU 1098 Ignatius's puzzle
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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".
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
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;}
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