Ignatius's puzzle
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 475 Accepted Submission(s): 269
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
#include<iostream>
using namespace std;
int main()
{
int k,i,sum;
while(cin>>k){
if(k%65==0) {
printf("no\n");
continue;
}
for(i=1;i<66;++i){
sum=i*k;
if((sum%65)==47) break;
}
if(i==66)
printf("no\n");
else
cout<<i<<endl;
}
return 0;
}题目的关键是函数式f(x)=5*x^13+13*x^5+k*a*x;
事实上，由于x取任何值都需要能被65整除.那么用数学归纳法.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.
那么把f(x+1)展开,得到5*( ( 13 0 )x^13 + (13 1 ) x^12 ...... .....+(13 13) x^0)+13*( ( 5 0 )x^5+(5 1 )x^4......其实就是二项式展开,这里就省略了 ......+ ( 5 5 )x^0 )+k*a*x+k*a;——————这里的( n m)表示组合数,相信学过2项式定理的朋友都能看明白.
然后提取出5*x^13+13*x^5+k*a*x
则f(x+1 ) = f (x) + 5*( (13 1 ) x^12 ...... .....+(13 13) x^0 )+ 13*( (5 1 )x^4+...........+ ( 5 5 )x^0 )+k*a;
很容易证明，除了5*(13 13) x^0 、13*( 5 5 )x^0 和k*a三项以外,其余各项都能被65整除.
那么也只要求出18+k*a能被65整除就可以了.
而f(1)也正好等于18+k*a
所以,只要找到a,使得18+k*a能被65整除,也就解决了这个题目.
此题属于纸老虎型^_^（如果没有想到这里就要动用暴力........默哀。）