Tonelli–Shanks Algorithm 二次剩余系解法 (Ural 1132. Square Root)

wikipedia上的解释和证明：http://en.wikipedia.org/wiki/Tonelli%E2%80%93Shanks_algorithm

The Tonelli–Shanks algorithm (referred to by Shanks as the RESSOL algorithm) is used within modular arithmetic to solve a congruence of the form

where n is a quadratic residue (mod p), and p is an odd prime.

Tonelli–Shanks cannot be used for composite moduli; finding square roots modulo composite numbers is a computational problem equivalent to integer factorization.^[1]

An equivalent, but slightly more redundant version of this algorithm was developed by Alberto Tonelli in 1891. The version discussed here was developed independently by Daniel Shanksin 1973, who explained:

"My tardiness in learning of these historical references was because I had lent Volume 1 of Dickson's History to a friend and it was never returned."^[2]

(Note: All  are taken to mean , unless indicated otherwise).[edit]The algorithm

Inputs: p, an odd prime. n, an integer which is a quadratic residue (mod p), meaning that the Legendre symbol .

Outputs: R, an integer satisfying .

1. Factor out powers of 2 from p − 1, defining Q and S as:  with Q odd. Note that if , i.e. , then solutions are given directly by .
2. Select a z such that the Legendre symbol  (that is, z should be a quadratic non-residue modulo p), and set .
3. Let 
4. Loop:
   1. If , return R.
   2. Otherwise, find the lowest i, , such that ; e.g. via repeated squaring.
   3. Let , and set  and .

Once you have solved the congruence with R the second solution is p − R.

Example

Solving the congruence . It is clear that  is odd, and since , 10 is a quadratic residue (by Euler's criterion).

• Step 1: Observe  so , .

• Step 2: Take  as the quadratic nonresidue (2 is a quadratic nonresidue since  (again, Euler's criterion)). Set 

• Step 3: 

• Step 4: Now we start the loop:  so ; i.e. 
  • Let , so .
  • Set . Set , and 
  • We restart the loop, and since  we are done, returning 

Indeed, observe that  and naturally also . So the algorithm yields two solutions to our congruence.

Proof

First write . Now write  and , observing that . This latter congruence will be true after every iteration of the algorithm's main loop. If at any point,  then  and the algorithm terminates with .

If , then consider , a quadratic non-residue of . Let . Then  and , which shows that the order of  is .

Similarly we have , so the order of  divides . Suppose the order of  is . Since  is a square modulo ,  is also a square, and hence .

Now we set  and with this ,  and . As before, holds; however with this construction both  and  have order . This implies that  has order  with .

If  then , and the algorithm stops, returning . Else, we restart the loop with analogous definitions of , , and  until we arrive at an  that equals 0. Since the sequence of S is strictly decreasing the algorithm terminates.

 

 

算法的大体过程已经说的很清楚，然后模拟那个过程就可以了，不过对于Ural上的这个题，要对n = 2特判一下。详见代码：

 

 

//#pragma comment(linker,"/STACK:327680000,327680000")
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>

#define CL(arr, val)    memset(arr, val, sizeof(arr))
#define REP(i, n)       for((i) = 0; (i) < (n); ++(i))
#define FOR(i, l, h)    for((i) = (l); (i) <= (h); ++(i))
#define FORD(i, h, l)   for((i) = (h); (i) >= (l); --(i))
#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d\n", x)
#define Read()  freopen("data.in", "r", stdin)
#define Write() freopen("data.out", "w", stdout);

typedef long long LL;
const double eps = 1e-8;
const double pi = acos(-1.0);
const double inf = ~0u>>2;


using namespace std;

LL MOD;

LL mod_exp(LL a, LL b) {
    LL res = 1;
    while(b > 0) {
        if(b&1)    res = (res*a)%MOD;
        a = (a*a)%MOD;
        b >>= 1;
    }
    return res;
}

LL solve(LL n, LL p) {
    int Q = p - 1, S = 0;
    while(Q%2 == 0) {
        Q >>= 1;
        S++;
    }
    if(S == 1) {return mod_exp(n, (p + 1)/4);}
    int z;
    while(1) {
        z = 1 + rand()%(p - 1);
        if(mod_exp(z, (p - 1)/2) != 1)   break;
    }
    LL c = mod_exp(z, Q);
    LL R = mod_exp(n, (Q + 1)/2);
    LL t = mod_exp(n, Q);
    LL M = S, b, i;
    while(1) {
        if(t%p == 1)  break;
        for(i = 1; i < M; ++i) {
            if(mod_exp(t, 1<<i) == 1)    break;
        }
        b = mod_exp(c, 1<<(M-i-1));
        R = (R*b)%p;
        t = (t*b*b)%p;
        c = (b*b)%p;
        M = i;
    }
    return (R%p + p)%p;
}

int main() {
    //Read();

    int T, a, n, i;
    scanf("%d", &T);
    while(T--) {
        scanf("%d%d", &a, &n);
        if(n == 2) {    //***
            if(a%n == 1)    printf("1\n");
            else    puts("No root");
            continue;
        }
        MOD = n;
        if(mod_exp(a, (n-1)/2) != 1)    {puts("No root"); continue; }

        i = solve(a, n);
        if(i == n - i)  printf("%d\n", i);
        else    printf("%d %d\n", min(i, n - i), max(i, n - i));
    }
    return 0;
}

 

 

 POJ 1808 勒让德符号(Legendre symbol)判定

 

View Code

//#pragma comment(linker,"/STACK:327680000,327680000")
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>

#define CL(arr, val)    memset(arr, val, sizeof(arr))
#define REP(i, n)       for((i) = 0; (i) < (n); ++(i))
#define FOR(i, l, h)    for((i) = (l); (i) <= (h); ++(i))
#define FORD(i, h, l)   for((i) = (h); (i) >= (l); --(i))
#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d\n", x)
#define Read()  freopen("data.in", "r", stdin)
#define Write() freopen("data.out", "w", stdout);

typedef long long LL;
const double eps = 1e-8;
const double pi = acos(-1.0);
const double inf = ~0u>>2;


using namespace std;

LL MOD;

LL mod_exp(LL a, LL b) {
    LL res = 1;
    while(b > 0) {
        if(b&1)    res = (res*a)%MOD;
        a = (a*a)%MOD;
        b >>= 1;
    }
    return res;
}

int main() {
    //Read();

    LL a, p;
    int T, cas = 0;
    scanf("%d", &T);
    while(T--) {
        scanf("%lld%lld", &a, &p);
        MOD = p;
        printf("Scenario #%d:\n", ++cas);
        if(mod_exp((a%p+p)%p, (p - 1)/2) == 1)  puts("1");    //注意a有可能是负数
        else    puts("-1");
        cout << endl;
    }
}