Mod Tree
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3543 Accepted Submission(s): 917
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
lcy
题目意思 k^x (mod P) = N
1.当N>=P,显然无解。
2.当P=1 ,显然为0。
为什么上一题poj3243那题 没有讨论这个呢。
上一题的意思是 A^x % C = B % C ,所以没有1的判断
这道题的输出很坑的，要特别注意。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cstring> 5 #include<cmath> 6 using namespace std; 7 const int MAX=499991; 8 typedef __int64 LL; 9 LL A,B,C; 10 bool Hash[MAX]; 11 LL val[MAX]; 12 LL idx[MAX]; 13 14 LL gcd(LL a,LL b) 15 { 16 if(b==0) 17 return a; 18 return gcd(b,a%b); 19 } 20 21 void Ex_gcd(LL a,LL b,LL &x,LL &y) 22 { 23 if(b==0) 24 { 25 x=1; 26 y=0; 27 return ; 28 } 29 Ex_gcd(b,a%b,x,y); 30 LL hxl=x-a/b*y; 31 x=y; 32 y=hxl; 33 return ; 34 } 35 36 void Insert(LL id,LL num)//哈希建立 37 { 38 LL k=num%MAX; 39 while(Hash[k] && val[k]!=num) 40 { 41 k++; if(k==MAX) k=k-MAX; 42 } 43 if(!Hash[k]) 44 { 45 Hash[k]=true; 46 val[k]=num; 47 idx[k]=id; 48 } 49 return; 50 } 51 52 LL found(LL num)//哈希查询 53 { 54 LL k=num%MAX; 55 while(Hash[k] && val[k]!=num) 56 { 57 k++; 58 if(k==MAX) k=k-MAX; 59 } 60 if(Hash[k]) 61 { 62 return idx[k]; 63 } 64 return -1; 65 } 66 67 LL baby_step(LL a,LL b,LL c) 68 { 69 LL temp=1; 70 for(LL i=0;i<=100;i++) 71 { 72 if(temp==b) return i; 73 temp=(temp*a)%c; 74 } 75 LL tmp,D=1,count=0; 76 memset(Hash,false,sizeof(Hash)); 77 memset(val,-1,sizeof(val)); 78 memset(idx,-1,sizeof(idx)); 79 80 while( (tmp=gcd(a,c))!=1 ) 81 { 82 if(b%tmp) return -1; 83 c=c/tmp; 84 b=b/tmp; 85 D=D*a/tmp%c; 86 count++; 87 } 88 LL cur=1; 89 LL M=ceil(sqrt(c*1.0)); 90 for(LL i=1;i<=M;i++)//初始的时候，i=0开始，错了。！！？ 91 { 92 cur=cur*a%c; 93 Insert(i,cur); 94 } 95 LL x,y; 96 for(LL i=0;i<=M;i++) 97 { 98 Ex_gcd(D,c,x,y); 99 x=x*b%c; 100 x=(x%c+c)%c; 101 LL k=found(x); 102 if(k!=-1) 103 { 104 return i*M+k+count; 105 } 106 D=D*cur%c; 107 } 108 return -1; 109 } 110 111 int main() 112 { 113 while(scanf("%I64d%I64d%I64d",&A,&C,&B)>0) 114 { 115 if(B>=C) 116 { 117 printf("Orz,I can’t find D! "); 118 continue; 119 } 120 if(C==1) 121 { 122 printf("0 "); 123 continue; 124 } 125 LL cur=baby_step(A,B,C); 126 if(cur==-1) 127 { 128 printf("Orz,I can’t find D! "); 129 } 130 else printf("%I64d ",cur); 131 } 132 return 0; 133 }