hdu1098 -数论

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Ignatius's puzzle

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6563 Accepted Submission(s): 4544

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

22no43

Author

eddy

思路：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。
那么把f(x+1)展开，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.

（参考原文）

即求18+k*a能被65整除的最小正数a，即求解18+k*a=65*n，利用扩展欧几里得算法求解即可

#include<iostream>#include<cmath>using namespace std;void extend_gcd(int x,int y,int &d,int &a,int &b){if(y==0){d=x;a=1;b=0;}else{extend_gcd(y,x%y,d,b,a);b-=a*(x/y);//a-=b*(x/y);}}int main(){int i,j,k,m,n,d,a;while(cin>>k){ extend_gcd(65,k,d,n,a);// cout<<d<<" "<<a<<" "<<n<<endl; if(18%d)cout<<"no\n"; else{ a*=18/d; if(a>0) a=a-(int)ceil((double)(a)/(65/d))*(65/d); else a=a+(-a)/(65/d)*(65/d); cout<<-a<<endl; }}return 0;}

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