HDUOJ-----1098 Ignatius's puzzle
Ignatius's puzzle
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5035 Accepted Submission(s): 3426
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
11 100 9999
Sample Output
22 no 43
Author
eddy
这道题，如果能够在高中的时候，很定能够求出来的..
对于 F(x)=5*x^13+13*x^5+k*a*x;
数学归纳如下：
当x=1时，F(1)=5+13+k*a (此时需要找一个a使得，条件成立)。
当x=1时，F(1)=5+13+k*a (此时需要找一个a使得，条件成立)。
假设 当x=k时，其成立；
当.x=k+1时，也应当成立。但实际上5*（x+1）^13+13*(x+1)^5+k*a*(x+1);
展开之后 0 13
5*（C *x^13.........C *x^0)+13*(.....同前头.....)+k*a*x + k*a ;
13 13
最后我5*x^13+13*x^5+kax +5+13+ka +后面部分（由于是65的倍数，就不讲啦）....由于当x=xs十，条件是成立的....所以最后只需要将（13+5+ka)%65
等于0就行了！！而a必须小于65,不然就没有意义了！！试想a如果是65的倍数的话....那么F(x+1)处，便放到后头省掉了...
由此就不难写出代码了！！
代码如下:
1 #include<iostream> 2 using namespace std; 3 int main( void ) 4 { 5 int k,a; 6 while(cin>>k) 7 { 8 a=0; 9 for(int i=1;i<65;i++) 10 { 11 if((18+i*k)%65==0) 12 { a=i; 13 break; 14 } 15 } 16 if(a)cout<<a<<endl; 17 else cout<<"no"<<endl; 18 19 } 20 return 0; 21 }