Discrete Logging
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
HinThe 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) .
题解：这道题目是Baby-step giant-step的裸题吧，嗯嗯，不要忘了开long long就可以了。
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<cstring>
#include<map>
#define ll long long
using namespace std; ll p,b,n; int ex_gcd(ll a,ll b,ll &x,ll &y)
{
if (!b){x=,y=;return a;}
else
{
int fzy=ex_gcd(b,a%b,x,y);
int t=x;x=y;
y=t-a/b*y;
return fzy;
}
}
void solve()
{
ll m=(ll)sqrt(p);
map<int,int>num;
map<int,bool>app;
app[]=,num[]=;
ll zhi=;
for (int i=;i<=m-;i++)
{
zhi=zhi*b%p;
if (!app[zhi])
{
app[zhi]=;
num[zhi]=i;
}
}
zhi=zhi*b%p;
ll x,y,nn=n;
int fzy=ex_gcd(zhi,p,x,y);
x=(x+p)%p;
for (int i=;i<=m;i++)
{
if (app[nn])
{
printf("%lld\n",i*m+num[nn]);
return;
}
else nn=nn*x%p;
}
printf("no solution\n");
}
int main()
{
while(~scanf("%d%d%d",&p,&b,&n))
solve();
}