poj2417

maksyuki 发表于 oj 分类，标签: 数论-离散对数

15 9月 2015

Discrete Logging

Given a prime P, 2 <= P < 2³¹, an integer B, 2 <= B < P, and an integer N, 1 <= N < P, compute the discrete logarithm of N, base B, modulo P. That is, find an integer L such that

B^L == N (mod P)

Input

Read several lines of input, each containing P,B,N separated by a space.

Output

For each line print the logarithm on a separate line. If there are several, print the smallest; if there is none, print "no solution".

Sample Input

5 2 1
5 2 2
5 2 3
5 2 4
5 3 1
5 3 2
5 3 3
5 3 4
5 4 1
5 4 2
5 4 3
5 4 4
12345701 2 1111111
1111111121 65537 1111111111

Sample Output

0
1
3
2
0
3
1
2
0
no solution
no solution
1
9584351
462803587

Hint

The solution to this problem requires a well known result in number theory that is probably expected of you for Putnam but not ACM competitions. It is Fermat's theorem that states

B^(P-1) == 1 (mod P)
for any prime P and some other (fairly rare) numbers known as base-B pseudoprimes. A rarer subset of the base-B pseudoprimes, known as Carmichael numbers, are pseudoprimes for every base between 2 and P-1. A corollary to Fermat's theorem is that for any m

B^(-m) == B^(P-1-m) (mod P) .

Source

Waterloo Local 2002.01.26

题目类型：形如 A ^ X = B (mod C)的高次同余方程

算法分析：直接按照A ^ X = B (mod C)的格式求解即可

#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>

#define lson rt << 1, l, m
#define rson rt << 1 | 1, m + 1, r

using namespace std;

const int INF = 0x7FFFFFFF;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int MOD = 7;
const int maxn = 110000 + 66;

struct HashNode
{
    long long data, id, next;
};

HashNode Hash[maxn<<1];
bool flag[maxn<<1];
long long top;

void Insert (long long a, long long b)
{
    long long k = b & maxn;
    if (flag[k] == false)
    {
        flag[k] = true;
        Hash[k].next = -1;
        Hash[k].id = a;
        Hash[k].data = b;
        return ;
    }

    while (Hash[k].next != -1)
    {
        if (Hash[k].data == b)
            return ;
        k = Hash[k].next;
    }

    if (Hash[k].data == b)
        return ;

    Hash[k].next = ++top;
    Hash[top].next = -1;
    Hash[top].id = a;
    Hash[top].data = b;
}

long long Find (long long b)
{
    long long k = b & maxn;
    if (flag[k] == false)
        return -1;

    while (k != -1)
    {
        if (Hash[k].data == b)
            return Hash[k].id;
        k = Hash[k].next;
    }

return -1;
}

long long gcd (long long a, long long b)
{
    return b ? gcd (b, a % b) : a;
}

long long gcd_ex (long long a, long long b, long long &x, long long &y)
{
    if (b == 0) { x = 1; y = 0; return a; }
    long long d = gcd_ex (b, a % b, y, x);
    y = y - a / b * x;
    return d;
}


long long q_mul_mod (long long a, long long b, long long p)
{
    long long ans = 1;
    while (b)
    {
        if (b & 1)
        {
            ans = ((ans % p) * (a % p)) % p;
            b--;
        }

        b = b >> 1;
        a = ((a % p) * (a % p)) % p;
    }

return ans;
}

long long BabyStep_GiantStep (long long A, long long B, long long C)
{
    top = maxn;  B %= C;
    long long tmp = 1, i;

    for (i = 0; i <= 100; tmp = tmp * A % C, i++)
        if (tmp == B % C)
            return i;

    long long D = 1, cnt = 0;
    while ((tmp = gcd(A, C)) != 1)
    {
        if(B % tmp)
            return -1;

        C /= tmp;
        B /= tmp;
        D = D * A / tmp % C;
        cnt++;
    }

    long long M = (long long) ceil (sqrt (C + 0.0));

    for (tmp = 1, i = 0; i <= M; tmp = tmp * A % C, i++)
        Insert (i, tmp);

    long long x, y, K = q_mul_mod (A, M, C);
    for (i = 0; i <= M; i++)
    {
        gcd_ex (D, C, x, y); // D * X = 1 ( mod C )
        tmp = ((B * x) % C + C) % C;

        if((y = Find(tmp)) != -1)
            return i * M + y + cnt;

        D = D * K % C;
    }

return -1;
}

int main()
{
//    ifstream cin ("aaa.txt");

    long long A, B, C;
    while(cin >> C >> A >> B)
    {
        memset (flag, 0, sizeof(flag));
        long long ans = BabyStep_GiantStep (A, B, C);

        if (ans == -1)
            cout << "no solution" << endl;

        else
            cout << ans << endl;
    }

return 0;
}

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#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

#define lson rt << 1, l, m

#define rson rt << 1 | 1, m + 1, r

using namespace std;

const int INF = 0x7FFFFFFF;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int MOD = 7;

const int maxn = 110000 + 66;

struct HashNode

{

long long data, id, next;

};

HashNode Hash[maxn<<1];

bool flag[maxn<<1];

long long top;

void Insert (long long a, long long b)

{

long long k = b & maxn;

if (flag[k] == false)

{

flag[k] = true;

Hash[k].next = -1;

Hash[k].id = a;

Hash[k].data = b;

return ;

}

while (Hash[k].next != -1)

{

if (Hash[k].data == b)

return ;

k = Hash[k].next;

}

if (Hash[k].data == b)

return ;

Hash[k].next = ++top;

Hash[top].next = -1;

Hash[top].id = a;

Hash[top].data = b;

}

long long Find (long long b)

{

long long k = b & maxn;

if (flag[k] == false)

return -1;

while (k != -1)

{

if (Hash[k].data == b)

return Hash[k].id;

k = Hash[k].next;

}

return -1;

}

long long gcd (long long a, long long b)

{

return b ? gcd (b, a % b) : a;

}

long long gcd_ex (long long a, long long b, long long &x, long long &y)

{

if (b == 0) { x = 1; y = 0; return a; }

long long d = gcd_ex (b, a % b, y, x);

y = y - a / b * x;

return d;

}

long long q_mul_mod (long long a, long long b, long long p)

{

long long ans = 1;

while (b)

{

if (b & 1)

{

ans = ((ans % p) * (a % p)) % p;

b--;

}

b = b >> 1;

a = ((a % p) * (a % p)) % p;

}

return ans;

}

long long BabyStep_GiantStep (long long A, long long B, long long C)

{

top = maxn; B %= C;

long long tmp = 1, i;

for (i = 0; i <= 100; tmp = tmp * A % C, i++)

if (tmp == B % C)

return i;

long long D = 1, cnt = 0;

while ((tmp = gcd(A, C)) != 1)

{

if(B % tmp)

return -1;

C /= tmp;

B /= tmp;

D = D * A / tmp % C;

cnt++;

}

long long M = (long long) ceil (sqrt (C + 0.0));

for (tmp = 1, i = 0; i <= M; tmp = tmp * A % C, i++)

Insert (i, tmp);

long long x, y, K = q_mul_mod (A, M, C);

for (i = 0; i <= M; i++)

{

gcd_ex (D, C, x, y); // D * X = 1 ( mod C )

tmp = ((B * x) % C + C) % C;

if((y = Find(tmp)) != -1)

return i * M + y + cnt;

D = D * K % C;

}

return -1;

}

int main()

{

// ifstream cin ("aaa.txt");

long long A, B, C;

while(cin >> C >> A >> B)

{

memset (flag, 0, sizeof(flag));

long long ans = BabyStep_GiantStep (A, B, C);

if (ans == -1)

cout << "no solution" << endl;

else

cout << ans << endl;

}

return 0;

}

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