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
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".
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) .
题解
1 //It is made by Awson on 2018.1.15 2 #include <set> 3 #include <map> 4 #include <cmath> 5 #include <ctime> 6 #include <queue> 7 #include <stack> 8 #include <cstdio> 9 #include <string> 10 #include <vector> 11 #include <cstdlib> 12 #include <cstring> 13 #include <iostream> 14 #include <algorithm> 15 #define LL long long 16 #define Abs(a) ((a) < 0 ? (-(a)) : (a)) 17 #define Max(a, b) ((a) > (b) ? (a) : (b)) 18 #define Min(a, b) ((a) < (b) ? (a) : (b)) 19 #define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b)) 20 using namespace std; 21 const LL MOD = 233333; 22 void read(LL &x) { 23 char ch; bool flag = 0; 24 for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar()); 25 for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar()); 26 x *= 1-2*flag; 27 } 28 void write(LL x) { 29 if (x > 9) write(x/10); 30 putchar(x%10+48); 31 } 32 33 LL a, b, c, ans; 34 struct MAP { 35 LL ha[MOD+5]; int id[MOD+5]; 36 void clear() {for (int i = 0; i < MOD; i++) ha[i] = id[i] = -1; } 37 int count(LL x) { 38 LL pos = x%MOD; 39 while (true) { 40 if (ha[pos] == -1) return 0; 41 if (ha[pos] == x) return 1; 42 ++pos; if (pos >= MOD) pos -= MOD; 43 } 44 } 45 void insert(LL x, int idex) { 46 LL pos = x%MOD; 47 while (true) { 48 if (ha[pos] == -1 || ha[pos] == x) {ha[pos] = x, id[pos] = idex; return; } 49 ++pos; if (pos >= MOD) pos -= MOD; 50 } 51 } 52 int query(LL x) { 53 LL pos = x%MOD; 54 while (true) { 55 if (ha[pos] == x) return id[pos]; 56 ++pos; if (pos >= MOD) pos -= MOD; 57 } 58 } 59 }mp; 60 61 LL quick_pow(LL a, LL b, LL c) { 62 LL ans = 1; 63 while (b) { 64 if (b&1) ans = ans*a%c; 65 a = a*a%c, b >>= 1; 66 } 67 return ans; 68 } 69 LL BSGS(LL a, LL b, LL c) { 70 mp.clear(); 71 LL tim = ceil(sqrt(c)), tmp = b%c; 72 for (int i = 0; i <= tim; i++) { 73 mp.insert(tmp, i); tmp = tmp*a%c; 74 } 75 LL t = tmp = quick_pow(a, tim, c); 76 for (int i = 1; i <= tim; i++) { 77 if (mp.count(tmp)) return tim*i-mp.query(tmp); 78 tmp = tmp*t%c; 79 } 80 return -1; 81 } 82 void work() { 83 while (~scanf("%lld%lld%lld", &c, &a, &b)) 84 if ((ans = BSGS(a, b, c)) == -1) printf("no solution "); 85 else write(ans), putchar(' '); 86 } 87 int main() { 88 work(); 89 return 0; 90 }