BZOJ 3239--Discrete Logging(BSGS)
3239: Discrete Logging
Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 635 Solved: 413
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Description
Given a prime P, 2 <= P < 231, an integer B, 2 <= B < P, and an integer N, 2 <= N < P, compute the discrete logarithm of N, base B, modulo P. That is, find an 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 the logarithm on a separate line. If there are several, print the smallest; if there is none, print "no solution".
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)
Sample Input
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
1
3
2
0
3
1
2
0
no solution
no solution
1
9584351
462803587
题目链接:
http://www.lydsy.com/JudgeOnline/problem.php?id=3239
Solution
BSGS的模板题。。。。
对于本题的做法。。一般是先设 L = i * e + j 或 L = i * e - j 。。。
e = ceil(sqrt(P))。。就是假如算出 sqrt(P)= 1.14 ,e就等于2,往大的取整
然后枚举 i 和 j 的值就能做到 O(sqrt(P))。。。
代码
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<map>
#define LL long long
using namespace std; LL P,B,N,e,now;
map<LL,int>mp;
LL pow(LL p,LL q){
LL s=1;
while(q){
if(q&1) s=s*p%P;
q>>=1;
p=p*p%P;
}
return s;
}
void solve(){
mp.clear();
if(N==1 && B>0){
printf("0\n");
return;
}
if( (!B) && (!N) ){printf("1\n");return;}
if(!B){printf("no solution\n");return;}
e=ceil(sqrt(P));
now=N%P;
for(int j=1;j<=e;j++){
now=now*B%P;
if(!mp[now]) mp[now]=j;
}
B=pow(B,e);
now=1;
for(int i=1;i<=e;i++){
now=now*B%P;
if(mp[now]>0){
N=e*i-mp[now];
printf("%lld\n",N);
return;
}
}
printf("no solution\n");
return;
}
int main(){
while(scanf("%lld%lld%lld",&P,&B,&N)!=EOF) solve();
return 0;
}
This passage is made by Iscream-2001.
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