POJ 2417: Discrete Logging BSGS

最新推荐文章于 2020-01-31 14:59:38 发布

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最新推荐文章于 2020-01-31 14:59:38 发布

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分类专栏： bsgs —————————数学

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Discrete Logging

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 7598		Accepted: 3208

Description

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) .

裸题注意开long long

#include<cmath>
#include<ctime>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<iomanip>
#include<vector>
#include<string>
#include<bitset>
#include<queue>
#include<map>
#include<set>
using namespace std;

typedef long long ll;

inline ll read()
{
	ll x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch<='9'&&ch>='0'){x=10*x+ch-'0';ch=getchar();}
	return x*f;
}
void print(ll x)
{if(x<0)putchar('-'),x=-x;if(x>=10)print(x/10);putchar(x%10+'0');}

ll P,B,N;

map<ll,ll> mp;

inline int qpow(ll x,ll y)
{
	ll res(1);
	while(y)
	{
		if(y&1) res=1ll*res*x%P;
		x=1ll*x*x%P;
		y>>=1;
	}
	return res;
}

void BSGS()
{
	B%=P;N%=P;
	if(!B && !N){puts("1");return ;}
	if(!B){puts("no solution");return ;}
	mp.clear();
	ll m=ceil(sqrt(P-1));
	for(int i=1;i<=m;++i)
		N=N*B%P,mp[N]=i;
	ll ine(1),tmp=qpow(B,m);
	for(int i=1,j;i<=m;++i)
	{
		ine=ine*tmp%P;
		if(j=mp[ine])
		{
			print((i*m-j)%(P-1));puts("");
			return ;
		}
	}
	puts("no solution");
}

int main()
{
	while(scanf("%lld%lld%lld",&P,&B,&N)>0)
		BSGS();
	return 0;
}