poj2417 Discrete Logging

最新推荐文章于 2020-01-31 14:59:38 发布

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最新推荐文章于 2020-01-31 14:59:38 发布

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分类专栏： ACM解题报告数论文章标签： logging integer numbers insert input each

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  Discrete Logging
 
 
  Time Limit: 5000MS   Memory Limit: 65536KB   64bit IO Format: %I64d & %I64u
 
[Submit]   [Go Back]   [Status]
 
  Description
  
   
    Given a prime P, 2 <= P < 2
    ³¹, 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) .

[Submit] [Go Back] [Status]

/*

跟上一题一样

题意：XY mod Z = K，给出x，z，k求y。（0 ≤ X, Z, K ≤ 109）其中Y (0 ≤ Y < Z).假如找不到合适的答案，请输出“No Solution”。
求解：mod z产生的集合最多有z个元素，这构成一个群。
分析：其时这个方法我还没搞明白，是从网上搞到的，这个模版还是比较稳当的
代码：*/
#include <iostream>
#include <string.h>
#include <math.h>
using namespace std;


const long long mod=800000;
const long long MAX=800000;

struct data
{
    long long  m,n,next;
}ln[MAX];
long long st[MAX],s;

void insert(long long m,long long n)
{
    long long  mm = m%mod;
    long long  pos = st[mm];
    while(pos != -1 && ln[pos].m != m)
        pos = ln[pos].next;
    if(pos == -1)
    {
        ln[s].m = m;
        ln[s].n = n;
        ln[s].next = st[mm];
        st[mm] = s++;
    }
    else ln[pos].n = n;
}

long long find(long long m)
{
    long long  mm=m%mod;
    long long pos=st[mm];
    while(pos!=-1)
    {
        if(ln[pos].m==m)
            return ln[pos].n;
        pos=ln[pos].next;
    }
    return -1;
}

long long MOD(long long b,long long n,long long mod)
{
    long long t=1;
    while(n)
    {
        if(n & ((__int64)1))
            t = (t*b)%mod;
        b = (b*b)%mod;
        n >>= ((__int64)1);
    }
    return t%mod;
}
long long solve(long long X,long long Z,long long K)//用来求X^Y mod Z = k 中的Y
{
        long long  m,i,j,b;
        X %= Z;
        K %= Z;
        s = 0;
        memset(st, -1, sizeof(st));
        m = (int)sqrt((double)Z);
        long long t = K;
        for(i=0; i <= m; i++)
        {
            insert(t, i);
            t *= X;
            t %= Z;
        }
        b = MOD(X, m, Z);
        t = 1;
        long long ans = -1;
        for(i=1; i <= m; i++)
        {
            t *= b;
            t %= Z;
            long long x = find(t);
            if(x != -1)
            {
                ans = i*m-x;
                break;
            }
        }
        if(ans != -1 && MOD(X, ans, Z) == K)
        return ans;
        else return -1;

}
int main()
{
    long long Z,X,K;
    long long power;
    while(cin>>Z>>X>>K)
    {
        if(!X && !Z && !K)cout<<0<<endl;
        power=solve(X,Z,K);
        if(power!=-1)cout<<power<<endl;
        else printf("no solution\n");
    }
    return 0;
}