高次同余方程（Baby-Step Giant-Step）

Discrete LogarithmProblem  网页链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=121425#problem/D 传送门：nefu

题目描述：

题目大意：

模板题，a^x=b(mod p) p为素数和不是素数都可以
代码实现：
#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#define MOD 76543
 
using namespace std;
inths[MOD],head[MOD],next[MOD],id[MOD],top;
void insert1(int x,int y)
{
    int k=x%MOD;
    hs[top]=x;
    id[top]=y;
    next[top]=head[k];
    head[k]=top++;
}
 
int find1(int x)
{
    int k=x%MOD;
    for(int i=head[k];i!=-1;i=next[i])
        if(hs[i]==x)
        return id[i];
    return -1;
}
 
int BSGS(int a,int b,int n)
{
    memset(head,-1,sizeof(head));
    top=1;
    if(b==1)
        return 0;
    int m=sqrt(n*1.0),j;
    long long x=1,p=1;
    for(int i=0;i<m;i++,p=p*a%n)
        insert1(p*b%n,i);
    for(long long i=m; ;i+=m)
    {
        if((j=find1(x=x*p%n))!=-1)
            return i-j;
        if(i>n)
            break;
    }
    return -1;
}
 
int main()
{
    int p,a,b;
    scanf("%d",&p);
    while(scanf("%d",&a)!=EOF)
    {
        if(a==0)
            break;
        else
        {
            scanf("%d",&b);
            b=b%p;
            int tmp=BSGS(a,b,p);
            if(tmp<0)
                puts("0");
            else printf("%d\n",tmp);
        }
    }
    return 0;
}
 
Discrete Logging

Time Limit: 5000MS	Memory Limit: 65536K
Total Submissions: 5015	Accepted: 2277

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, findan 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 thelogarithm on a separate line. If there are several, print the smallest; ifthere 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 problemrequires a well known result in number theory that is probably expected of youfor 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-Bpseudoprimes. A rarer subset of the base-B pseudoprimes, known as Carmichaelnumbers, are pseudoprimes for every base between 2 and P-1. A corollary toFermat's theorem is that for any m 
  B(-m) == B(P-1-m) (mod P) .
Source
Waterloo Local 2002.01.26
 
题目大意：
同上，只需要对输入和输出稍作修改即可。
代码实现：
#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#define MOD 76543
 
using namespace std;
inths[MOD],head[MOD],next[MOD],id[MOD],top;
void insert1(int x,int y)
{
    int k=x%MOD;
    hs[top]=x;
    id[top]=y;
    next[top]=head[k];
    head[k]=top++;
}
 
int find1(int x)
{
    int k=x%MOD;
    for(int i=head[k];i!=-1;i=next[i])
        if(hs[i]==x)
        return id[i];
    return -1;
}
 
int BSGS(int a,int b,int n)
{
    memset(head,-1,sizeof(head));
    top=1;
    if(b==1)
        return 0;
    int m=sqrt(n*1.0),j;
    long long x=1,p=1;
    for(int i=0;i<m;i++,p=p*a%n)
        insert1(p*b%n,i);
    for(long long i=m; ;i+=m)
    {
        if((j=find1(x=x*p%n))!=-1)
            return i-j;
        if(i>n)
            break;
    }
    return -1;
}
 
int main()
{
    int p,a,b;
   while(scanf("%d%d%d",&p,&a,&b)!=EOF)
    {
        if(a==0&&p==0&&b==0)
            break;
        else
        {
            b=b%p;
            int tmp=BSGS(a,b,p);
            if(tmp<0)
                puts("no solution");
            else printf("%d\n",tmp);
        }
    }
    return 0;
}
 
Clever Y

Time Limit: 5000MS	Memory Limit: 65536K
Total Submissions: 7919	Accepted: 1971

Description
Little Y finds there is a veryinteresting 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 morethan 20 test cases. For each test case, there would be only one line containing3 integers X, Z, K (0 ≤ X, Z, K ≤ 109). 

Input file ends with 3 zeros separated by spaces. 
Output
For each test case output oneline. Write "No Solution" (without quotes) if you cannot find afeasible 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
 
题目大意：
同上，只需对输入输出稍作修改即可！
代码实现：
#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#define MOD 76543
 
using namespace std;
inths[MOD],head[MOD],next[MOD],id[MOD],top;
void insert1(int x,int y)
{
    int k=x%MOD;
    hs[top]=x;
    id[top]=y;
    next[top]=head[k];
    head[k]=top++;
}
 
int find1(int x)
{
    int k=x%MOD;
    for(int i=head[k];i!=-1;i=next[i])
        if(hs[i]==x)
        return id[i];
    return -1;
}
 
int BSGS(int a,int b,int n)
{
    memset(head,-1,sizeof(head));
    top=1;
    if(b==1)
        return 0;
    int m=sqrt(n*1.0),j;
    long long x=1,p=1;
    for(int i=0;i<m;i++,p=p*a%n)
        insert1(p*b%n,i);
    for(long long i=m; ;i+=m)
    {
        if((j=find1(x=x*p%n))!=-1)
            return i-j;
        if(i>n)
            break;
    }
    return -1;
}
 
int main()
{
    int p,a,b;
   while(scanf("%d%d%d",&a,&p,&b)!=EOF)
    {
        if(a==0&&p==0&&b==0)
            break;
        else
        {
            b=b%p;
            int tmp=BSGS(a,b,p);
            if(tmp<0)
                puts("No Solution");
            else printf("%d\n",tmp);
        }
    }
    return 0;
}