Discrete Logging
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) . Source |
模板题。
http://hi.baidu.com/aekdycoin/item/236937318413c680c2cf29d4
1 /* *********************************************** 2 Author :kuangbin 3 Created Time :2013/8/24 0:06:54 4 File Name :F:\2013ACM练习\专题学习\数学\Baby_step_giant_step\POJ2417.cpp 5 ************************************************ */ 6 7 #include <stdio.h> 8 #include <string.h> 9 #include <iostream> 10 #include <algorithm> 11 #include <vector> 12 #include <queue> 13 #include <set> 14 #include <map> 15 #include <string> 16 #include <math.h> 17 #include <stdlib.h> 18 #include <time.h> 19 using namespace std; 20 //baby_step giant_step 21 // a^x = b (mod n) n为素数，a,b < n 22 // 求解上式 0<=x < n的解 23 #define MOD 76543 24 int hs[MOD],head[MOD],next[MOD],id[MOD],top; 25 void insert(int x,int y) 26 { 27 int k = x%MOD; 28 hs[top] = x, id[top] = y, next[top] = head[k], head[k] = top++; 29 } 30 int find(int x) 31 { 32 int k = x%MOD; 33 for(int i = head[k]; i != -1; i = next[i]) 34 if(hs[i] == x) 35 return id[i]; 36 return -1; 37 } 38 int BSGS(int a,int b,int n) 39 { 40 memset(head,-1,sizeof(head)); 41 top = 1; 42 if(b == 1)return 0; 43 int m = sqrt(n*1.0), j; 44 long long x = 1, p = 1; 45 for(int i = 0; i < m; ++i, p = p*a%n)insert(p*b%n,i); 46 for(long long i = m; ;i += m) 47 { 48 if( (j = find(x = x*p%n)) != -1 )return i-j; 49 if(i > n)break; 50 } 51 return -1; 52 } 53 int main() 54 { 55 //freopen("in.txt","r",stdin); 56 //freopen("out.txt","w",stdout); 57 int a,b,n; 58 while(scanf("%d%d%d",&n,&a,&b) == 3) 59 { 60 int ans = BSGS(a,b,n); 61 if(ans == -1)printf("no solution\n"); 62 else printf("%d\n",ans); 63 } 64 return 0; 65 }
