Problem Description
The picture indicates a tree, every node has 2 children.
The depth of the nodes whose color is blue is 3; the depth of the node whose color is pink is 0.
Now out problem is so easy, give you a tree that every nodes have K children, you are expected to calculate the minimize depth D so that the number of nodes whose depth is D equals to N after mod P.
Input
The input consists of several test cases.
Every cases have only three integers indicating K, P, N. (1<=K, P, N<=10^9)
Every cases have only three integers indicating K, P, N. (1<=K, P, N<=10^9)
Output
The minimize D.
If you can’t find such D, just output “Orz,I can’t find D!”
If you can’t find such D, just output “Orz,I can’t find D!”
Sample Input
3 78992 453
4 1314520 65536
5 1234 67
Sample Output
Orz,I can’t find D!
8
20
Author
AekdyCoin
Source
Recommend
【分析】
普通的BSGS是只能够解决A^x = B(mod C),其中C为素数的形式，而通过网上的消元的方法能够解决这种问题。
同样BSGS直接解决C为素数的形式也是可以的.
1 /* 2 宋代晏几道 3 《生查子·狂花顷刻香》 4 狂花顷刻香，晚蝶缠绵意。天与短因缘，聚散常容易。 5 传唱入离声，恼乱双蛾翠。游子不堪闻，正是衷肠事。 6 */ 7 #include <iostream> 8 #include <cstdio> 9 #include <ctime> 10 #include <cmath> 11 #include <algorithm> 12 #include <cstring> 13 #include <string> 14 #include <map> 15 #include <set> 16 #include <vector> 17 #define LOCAL 18 const int MAXN = 1000 + 10; 19 const int INF = 0x7fffffff; 20 using namespace std; 21 typedef long long LL; 22 23 int gcd(int a, int b ){return b == 0 ? a : gcd(b, a % b);} 24 int ext_gcd(int a, int b, int &x, int &y){ 25 if (b == 0) {x = 1; y = 0; return a;} 26 int tmp = ext_gcd(b, a % b, y, x); 27 y -= x * (a / b); 28 return tmp; 29 } 30 //求解 31 int Inval(int a, int b, int n){ 32 int x, y, e; 33 ext_gcd(a, n, x, y); 34 e = (LL) x * b % n;//小心超出int 的范围,因为a,n是互质的，因此求解出来的结果就是ax + ny = 1，乘以b才正确的答案 35 return e < 0 ? e + n : e; 36 } 37 // k s m 38 int pow_mod(LL a, int b, int c){ 39 if (b == 1) return a % c; 40 LL tmp = pow_mod(a, b / 2, c); 41 if (b % 2 == 0) return (tmp * tmp) % c; 42 else return (((tmp * tmp) % c) * a) % c; 43 } 44 int BSGS(int A, int B, int C){ 45 map<int, int> H;//hash 46 LL buf = 1 % C, D = buf, K; 47 int d = 0, tmp; 48 //小步 49 for (int i = 0; i <= 100; buf = buf * A % C, i++) 50 if (buf == B) return i; 51 //消因子 52 while ((tmp = gcd(A, C)) != 1){ 53 if (B % gcd(A, C) != 0) return -1;//为了解不定方程 54 d++; 55 C /= tmp; 56 B /= tmp; 57 D = D * A / tmp % C; 58 } 59 H.clear(); 60 int M = (int)ceil(sqrt(C * 1.0)); 61 buf = 1 % C; 62 for (int i = 0; i <= M; buf = buf * A % C, i++) 63 if (H.find((int)buf) == H.end()) H[(int)buf] = i;//Hash 64 65 K = pow_mod ((LL) A, M, C); 66 for (int i = 0; i <= M; D = D * K % C, i++){ 67 tmp = Inval((int) D ,B, C);//D和C是互质的 68 //一定不要忘了最后的d 69 if (tmp >= 0 && H.find(tmp) != H.end()) return i * M + H[tmp] + d; 70 } 71 return -1;//找不到 72 } 73 int main(){ 74 75 //转换为A^x = B(mod C)的形式 76 int A, B, C; 77 while (scanf("%d%d%d", &A, &C, &B) != EOF){ 78 if (B >= C) {printf("Orz,I can’t find D! "); continue;}//233 79 int tmp = BSGS(A, B, C); 80 if (tmp < 0) printf("Orz,I can’t find D! "); 81 else printf("%d ", tmp); 82 } 83 return 0; 84 } 85 86