【二次剩余-欧拉准则】HDOJ Jacobi symbol 3589

Jacobi symbol Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 549    Accepted Submission(s): 207 Problem Description Consider a prime number p and an integer a !≡ 0 (mod p). Then a is called a quadratic residue mod p if there is an integer x such that x2 ≡ a (mod p), and a quadratic non residue otherwise. Lagrange introduced the following notation, called the Legendre symbol, L (a,p):  For the calculation of these symbol there are the following rules, valid only for distinct odd prime numbers p, q and integers a, b not divisible by p:  The Jacobi symbol, J (a, n) ,is a generalization of the Legendre symbol ,L (a, p).It defines as : 1.  J (a, n) is only defined when n is an odd. 2.  J (0, n) = 0. 3.  If n is a prime number, J (a, n) = L(a, n). 4.  If n is not a prime number, J (a, n) = J (a, p1) *J (a, p2)…* J (a, pm), p1…pm is the prime factor of n.   Input Two integer a and n, 2 < a< =106，2 < n < =106，n is an odd number.   Output Output J (a,n)   Sample Input 3 5 3 9 3 13   Sample Output -1 0 1   Author alpc41   Source 2010 ACM-ICPC Multi-University Training Contest（15）——Host by NUDT   Recommend zhouzeyong   |   We have carefully selected several similar problems for you:  3583 3590 3588 3587 3586

题意：

求J(a,n)

解题思路：

二次剩余，欧拉准则！

在数论中，特别在同余理论裏，一个整数 

 对另一个整数 

 的二次剩余（英语：Quadratic
residue）指 

 的平方 

 除以 

 得到的余数。

当对于某个

及某个

，式子 

 成立时，称“

是模

的二次剩余”

当对于某个

及某个

，

 不成立时，称“

是模

的二次非剩余”
  欧拉准则：

若

是奇质数且

不能整除

，则：


是模

的二次剩余当且仅当：




是模

的非二次剩余当且仅当：



以勒让德符号表示，即为： 


AC代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>

using namespace std;

const int MAXN = 1000000+10;
typedef long long LL;

bool is_prime[MAXN];
int prime[MAXN];

void init()
{
int tot=0;
is_prime[0]=is_prime[1]=true;
for(int i=2;i<MAXN;i++){
if(!is_prime[i]){
prime[tot++]=i;
for(int j=i+i;j<MAXN;j+=i){
is_prime[j]=true;
}
}
}
}

int fast_power(int a,int n,int M)
{
LL res=1;
LL x=a;
while(n){
if(n&1) res=res*x%M;
x=x*x%M;
n>>=1;
}
return res%M;
}

int solve(int a,int p)
{
if(a%p==0) return 0;
return fast_power(a,(p-1)/2,p)==1?1:-1;
}

int main ()
{
int a,n;
init();
while(scanf("%d%d",&a,&n)!=EOF){
int ans=1;
if(is_prime
){
for(int i=0;prime[i]<=n;i++){
if(n%prime[i]) continue;
int cnt=0;
while(n%prime[i]==0){
n/=prime[i];
cnt++;
}
int t=solve(a,prime[i]);
if(t==-1&&cnt%2==0) t=1;
ans*=t;
}
}
else ans=solve(a,n);
printf("%d\n",ans);
}
return 0;
}