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) .
计算离散对数模版题。
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
/*************************************
*baby_step giant_step
*应用条件n必须为素数
*a^x = b (mod n) ，a,b < n
*求解上式 0<=x < n的解
*************************************/
const int MOD = 76543;
int hs[MOD],head[MOD],Next[MOD],id[MOD],top;
void Insert(int x,int y)
{
int k = x%MOD;
hs[top] = x, id[top] = y, Next[top] = head[k], head[k] = top++;
}
int Find(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)Insert(p*b%n,i);
for(long long i = m; ;i += m)
{
if( (j = Find(x = x*p%n)) != -1 )return i-j;//找到了可行解
if(i > n)break;
}
//如果无解
return -1;
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int a,b,n;
while(scanf("%d%d%d",&n,&a,&b) == 3)
{
int ans = BSGS(a,b,n);
if(ans == -1)printf("no solution\n");
else printf("%d\n",ans);
}
return 0;
}
