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
大意：一个k叉数，求小深度D，使得叶子节点数模P为N
即求解kD≡N (mod P)
拓展BSGS
注意N不能大于等于P
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<cmath> 6 using namespace std; 7 typedef long long lol; 8 int MOD=250000; 9 lol ha[300001],id[300001]; 10 void insert(lol x,lol d) 11 { 12 lol pos=x%MOD; 13 while (1) 14 { 15 if (ha[pos]==-1||ha[pos]==x) 16 { 17 ha[pos]=x; 18 id[pos]=d; 19 return; 20 } 21 pos++; 22 if (pos>=MOD) pos-=MOD; 23 } 24 } 25 bool count(lol x) 26 { 27 lol pos=x%MOD; 28 while (1) 29 { 30 if (ha[pos]==-1) return 0; 31 if (ha[pos]==x) return 1; 32 pos++; 33 if (pos>=MOD) pos-=MOD; 34 } 35 } 36 lol query(lol x) 37 { 38 lol pos=x%MOD; 39 while (1) 40 { 41 if (ha[pos]==x) return id[pos]; 42 pos++; 43 if (pos>=MOD) pos-=MOD; 44 } 45 } 46 lol qpow(lol x,lol y,lol Mod) 47 { 48 lol res=1; 49 while (y) 50 { 51 if (y&1) res=res*x%Mod; 52 x=x*x%Mod; 53 y>>=1; 54 } 55 return res; 56 } 57 lol gcd(lol a,lol b) 58 { 59 if (!b) return a; 60 return gcd(b,a%b); 61 } 62 lol BSGS(lol a,lol b,lol Mod) 63 {lol i; 64 if (b==1) return 0; 65 if (a==0&&b!=0) return -1; 66 if (b>=Mod) return -1; 67 memset(ha,-1,sizeof(ha)); 68 memset(id,0,sizeof(id)); 69 lol cnt=0,d=1,t; 70 while ((t=gcd(a,Mod))!=1) 71 { 72 if (b%t) return -1; 73 cnt++; 74 b/=t;Mod/=t; 75 d=d*(a/t)%Mod; 76 if (d==b) return cnt; 77 } 78 lol tim=sqrt((double)Mod); 79 lol tmp=b%Mod; 80 for (i=0;i<=tim;i++) 81 { 82 insert(tmp,i); 83 tmp=tmp*a%Mod; 84 } 85 t=tmp=qpow(a,tim,Mod); 86 tmp=tmp*d%Mod; 87 for (i=1;i<=tim;i++) 88 { 89 if (count(tmp)) 90 return i*tim-query(tmp)+cnt; 91 tmp=tmp*t%Mod; 92 } 93 return -1; 94 } 95 int main() 96 {lol a,b,Mod,ans; 97 while (cin>>a>>Mod>>b) 98 { 99 ans=BSGS(a,b,Mod); 100 if (ans==-1) printf("Orz,I can’t find D! "); 101 else printf("%lld ",ans); 102 } 103 }