【poj2417】Discrete Logging-优快云博客

Time Limit: 5000MS Memory Limit: 65536K

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

Solution

一道BSGS模板题啦，留作纪念……

Code

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long ll;
const ll mod=76543;
ll P,B,N,ans,top,hs[mod],id[mod],Last[mod],Next[mod];
void Insert(ll x,ll y)
{
    ll z=x%mod;
    hs[++top]=x,id[top]=y,Next[top]=Last[z],Last[z]=top;
}
ll Find(ll x)
{
    ll z=x%mod;
    for(ll i=Last[z];i;i=Next[i])
        if(hs[i]==x) return id[i];
    return -1;
}
ll BSGS(ll a,ll b,ll c)
{
    if(b==1) return 0;
    ll m=ceil(sqrt(c)),j=1,t=1,k;
    for(ll i=0;i<m;i++,j=j*a%c) Insert(j*b%c,i);
    for(ll i=m;i<=c;i+=m)
        if(~(k=Find(t=t*j%c))) return i-k;
    return -1;
}
int main()
{
    while(~scanf("%lld%lld%lld",&P,&B,&N))
    {
        top=0,memset(Last,0,sizeof(Last));
        ans=BSGS(B,N,P);
        if(~ans) printf("%lld\n",ans);
        else puts("no solution");
    }
    return 0;
}