【POJ 3243-Clever Y】 与【POJ 2417-Discrete Logging】（解高次同余方程 Baby-Step-Gaint-Step）

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:

XY 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

题目意思：

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

解题思路：

模板题，解高次同余方程。

#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 

BL == 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//这里写大于65535的数会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**/