Mod Tree(hdu2815)

Mod Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5934    Accepted Submission(s): 1498

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)

 

Output

The minimize 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

扩展baby_step,giant_step算法模板题

  1 #include<stdio.h>
  2 #include<algorithm>
  3 #include<queue>
  4 #include<stack>
  5 #include<string.h>
  6 #include<iostream>
  7 #include<math.h>
  8 #include<map>
  9 using namespace std;
 10 typedef long long LL;
 11 typedef struct node
 12 {
 13     LL val;
 14     int id;
 15 } ss;
 16 LL quick(LL n,LL m,LL mod);
 17 pair<LL,LL>ex_gcd(LL n,LL m);
 18 LL gcd(LL n,LL m);
 19 bool cmp(node p,node q);
 20 LL g_step_b_step(LL x,LL k,LL z);
 21 ss ans[100000];
 22 int main(void)
 23 {
 24     LL x,z,k;
 25     while(scanf("%lld %lld %lld",&x,&z,&k)!=EOF)
 26     {
 27             LL ask = g_step_b_step(x,k,z);
 28             if(ask == -1||k >= z)
 29                 printf("Orz,I can’t find D!\n");
 30             else printf("%lld\n",ask);
 31     }
 32     return 0;
 33 }
 34 LL g_step_b_step(LL x,LL k,LL z)
 35 {
 36     LL y = 0;
 37     LL xx = 1;
 38     while(true)
 39     {
 40         LL c = xx%z;
 41         if(c == k)return y;
 42         LL gc = gcd(x,z);
 43         if(gc == 1)break;
 44         y++;if(k%gc)return -1;
 45         z/=gc;k /= gc;
 46         xx = xx*(x/gc);
 47         xx%=z;
 48     }
 49     LL zz = sqrt(z) + 1;
 50     pair<LL,LL>NI = ex_gcd(x,z);
 51     NI.first = (NI.first%z + z)%z;
 52     LL NNI = NI.first*(k%z)%z;
 53     ans[0].id = 0,ans[0].val = k;
 54     for(int i = 1; i <= zz; i++)
 55     {
 56         ans[i].id = i;
 57         ans[i].val = NNI;
 58         NNI = NNI*NI.first%z;
 59     }
 60     sort(ans,ans+zz+1,cmp);
 61     LL x1 = quick(x,zz,z);
 62     LL slx = xx;
 63     for(int i = 0; i <= zz; i++)
 64     {
 65         int l = 0,r = zz;
 66         int id = -1;
 67         while(l <= r)
 68         {
 69             int mid = (l+r)/2;
 70             if(ans[mid].val >= slx)
 71             {
 72                 id = mid;
 73                 r = mid - 1;
 74             }
 75             else l = mid + 1;
 76         }
 77         if(id!=-1)
 78         {
 79             if(ans[id].val == slx)
 80             {
 81                 LL ask = (LL)i*zz + ans[id].id + y;
 82                 return ask;
 83             }
 84         }
 85         slx = slx*x1%z;
 86     }
 87     return -1;
 88 }
 89 LL gcd(LL n,LL m)
 90 {
 91     if(m == 0)
 92         return n;
 93     else return gcd(m,n%m);
 94 }
 95 LL quick(LL n,LL m,LL mod)
 96 {
 97     n%=mod;
 98     LL ask = 1;
 99     while(m)
100     {
101         if(m&1)
102             ask = ask*n%mod;
103         n = n*n%mod;
104         m/=2;
105     }
106     return ask;
107 }
108 pair<LL,LL>ex_gcd(LL n,LL m)
109 {
110     if(m == 0)
111         return make_pair(1,0);
112     else
113     {
114         pair<LL,LL>ans = ex_gcd(m,n%m);
115         return make_pair(ans.second,ans.first - (n/m)*ans.second);
116     }
117 }
118 bool cmp(node p,node q)
119 {
120     if(p.val == q.val)
121         return p.id < q.id;
122     else return p.val < q.val;
123 }