【POJ 3243-Clever Y】與【POJ 2417-Discrete Logging】（解高次同餘方程 Baby-Step-Gaint-Step）

kewlgrl發表於2016-08-03

Clever Y

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 7969		Accepted: 1986

Description

Little Y finds there is a very interesting formula in mathematics:

X^Y mod Z = K

Given X, Y, Z, we all know how to figure out K fast. However, given X, Z, K, could you figure out Y fast?

Input

Input data consists of no more than 20 test cases. For each test case, there would be only one line containing 3 integers X, Z, K (0 ≤ X, Z, K ≤ 10⁹).
Input file ends with 3 zeros separated by spaces.

Output

For each test case output one line. Write "No Solution" (without quotes) if you cannot find a feasible Y (0 ≤ Y < Z). Otherwise output the minimum Y you find.

Sample Input

5 58 33
2 4 3
0 0 0

Sample Output

9
No Solution

Source

POJ Monthly--2007.07.08, Guo, Huayang

題目意思：

已知X，Z和K，求解Y滿足： (X^Y) mod Z=K，即 (X^Y)≡K(mod Z)。

解題思路：

模板題，解高次同餘方程。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<malloc.h>
using namespace std;
typedef long long ll;
#define maxn 100000

struct Hash
{
    int a,b,next;
} hash[2*maxn];
int flag[maxn];
int top,idx;

void ins(int a,int b)
{
    int k=b&maxn;
    if(flag[k]!=idx)
    {
        flag[k]=idx;
        hash[k].next=-1;
        hash[k].a=a;
        hash[k].b=b;
        return;
    }
    while(hash[k].next!=-1)
    {
        if(hash[k].b==b) return;
        k=hash[k].next;
    }
    hash[k].next=++top;
    hash[top].next=-1;
    hash[top].a=a;
    hash[top].b=b;
}

int find(int b)
{
    int k=b&maxn;
    if(flag[k]!=idx) return -1;
    while(k!=-1)
    {
        if(hash[k].b==b) return hash[k].a;
        k=hash[k].next;
    }
    return -1;
}

int gcd(int a,int b)
{
    return b==0?a:gcd(b,a%b);
}

int exgcd(int a,int b,int &x,int &y)
{
    int t,d;
    if(!b)
    {
        x=1,y=0;
        return a;
    }
    d=exgcd(b,a%b,x,y);
    t=x,x=y,y=t-a/b*y;
    return d;
}

int inval(int a,int b,int n)
{
    int x,y,e;
    exgcd(a,n,x,y);
    e=(long long)x*b%n;
    return e<0?e+n:e;
}

int pow_mod(long long a,int b,int c)
{
    long long ret=1%c;
    a%=c;
    while(b)
    {
        if(b&1) ret=ret*a%c;
        a=a*a%c;
        b>>=1;
    }
    return ret;
}

int BabyStep(int A,int B,int C)
{
    top=maxn;
    ++idx;
    long long buf=1%C,D=buf,K;
    int i,d=0,tmp;
    for(i=0; i<=100; buf=buf*A%C,++i)
        if(buf==B) return i;
    while((tmp=gcd(A,C))!=1)
    {
        if(B%tmp) return -1;
        ++d;
        C/=tmp;
        B/=tmp;
        D=D*A/tmp%C;
    }
    int M=(int)ceil(sqrt((double)C));
    for(buf=1%C,i=0; i<=M; buf=buf*A%C,++i)
        ins(i,buf);
    for(i=0,K=pow_mod((long long)A,M,C); i<=M; D=D*K%C,++i)
    {
        tmp=inval((int)D,B,C);
        int w;
        if(tmp>=0&&(w=find(tmp))!=-1)
            return i*M+w+d;
    }
    return -1;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int A,B,C;
    while(cin>>A>>C>>B,A||B||C)
    {
        B%=C;
        int tmp=BabyStep(A,B,C);
        if(tmp<0) puts("No Solution");
        else cout<<tmp<<endl;
    }
    return 0;
}
/**
5 58 33
2 4 3
0 0 0
**/

Discrete Logging

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 5064		Accepted: 2301

Description

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

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<malloc.h>
using namespace std;
typedef long long ll;
#define maxn 65535//這裡寫大於65535的數會WA

struct Hash
{
    ll a,b,next;
} hash[2*maxn];
ll flag[maxn];
ll top,idx;

void ins(ll a,ll b)
{
    ll k=b&maxn;
    if(flag[k]!=idx)
    {
        flag[k]=idx;
        hash[k].next=-1;
        hash[k].a=a;
        hash[k].b=b;
        return;
    }
    while(hash[k].next!=-1)
    {
        if(hash[k].b==b) return;
        k=hash[k].next;
    }
    hash[k].next=++top;
    hash[top].next=-1;
    hash[top].a=a;
    hash[top].b=b;
}

ll find(ll b)
{
    ll k=b&maxn;
    if(flag[k]!=idx) return -1;
    while(k!=-1)
    {
        if(hash[k].b==b) return hash[k].a;
        k=hash[k].next;
    }
    return -1;
}

ll gcd(ll a,ll b)
{
    return b==0?a:gcd(b,a%b);
}

ll exgcd(ll a,ll b,ll &x,ll &y)
{
    ll t,d;
    if(!b)
    {
        x=1,y=0;
        return a;
    }
    d=exgcd(b,a%b,x,y);
    t=x,x=y,y=t-a/b*y;
    return d;
}

ll inval(ll a,ll b,ll n)
{
    ll x,y,e;
    exgcd(a,n,x,y);
    e=(long long)x*b%n;
    return e<0?e+n:e;
}

ll pow_mod(long long a,ll b,ll c)
{
    long long ret=1%c;
    a%=c;
    while(b)
    {
        if(b&1) ret=ret*a%c;
        a=a*a%c;
        b>>=1;
    }
    return ret;
}

ll BabyStep(ll A,ll B,ll C)
{
    top=maxn;
    ++idx;
    long long buf=1%C,D=buf,K;
    ll i,d=0,tmp;
    for(i=0; i<=100; buf=buf*A%C,++i)
        if(buf==B) return i;
    while((tmp=gcd(A,C))!=1)
    {
        if(B%tmp) return -1;
        ++d;
        C/=tmp;
        B/=tmp;
        D=D*A/tmp%C;
    }
    ll M=(ll)ceil(sqrt((double)C));
    for(buf=1%C,i=0; i<=M; buf=buf*A%C,++i)
        ins(i,buf);
    for(i=0,K=pow_mod((long long)A,M,C); i<=M; D=D*K%C,++i)
    {
        tmp=inval((ll)D,B,C);
        ll w;
        if(tmp>=0&&(w=find(tmp))!=-1)
            return i*M+w+d;
    }
    return -1;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    ll A,B,C;
    while(cin>>C>>A>>B)
    {
        B%=C;
        ll tmp=BabyStep(A,B,C);
        if(tmp<0) puts("no solution");
        else cout<<tmp<<endl;
    }
    return 0;
}

/**
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
**/

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【POJ 3243-Clever Y】 與【POJ 2417-Discrete Logging】（解高次同餘方程 Baby-Step-Gaint-Step）

題目意思：

解題思路：

相關文章

【POJ 3243-Clever Y】與【POJ 2417-Discrete Logging】（解高次同餘方程 Baby-Step-Gaint-Step）