hdu 2815 Mod Tree （扩展BSGS）

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Mod Tree

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6172 Accepted Submission(s): 1544

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 4534 1314520 655365 1234 67

Sample Output

Orz,I can’t find D!820

Author

AekdyCoin

Source

HDU 1st “Old-Vegetable-Birds Cup” Programming Open Contest

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题解：扩展BSGS

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<map>#define LL long long using namespace std;LL a,p,b;map<LL,LL> mp;LL quickpow(LL num,LL x){LL base=num%p; LL ans=1;while (x) {if (x&1) ans=ans*base%p;x>>=1;    base=base*base%p;}return ans;}LL gcd(LL x,LL y){LL r;while (y) {r=x%y;x=y; y=r;}return x;}LL exbsgs(LL a,LL b,LL p){a%=p; b%=p;if (b==1) return 0;LL cnt=0,d=1,tmp=1;while ((tmp=gcd(a,p))!=1) {if (b%tmp) return -1;b/=tmp; p/=tmp; cnt++;d=d*(a/tmp)%p;if (b==d) return cnt;}LL m=ceil(sqrt(p)); LL ans=b; LL sum=1;mp.clear();tmp=quickpow(a,m);mp[ans]=0;for (LL i=1;i<=m;i++) ans=ans*a%p,mp[ans]=i;    for (int LL i=1;i<=m+1;i++) {    d=d*tmp%p;    if (mp[d]) {    return i*m-mp[d]+cnt;}}return -1;}int main(){freopen("a.in","r",stdin);while (scanf("%I64d%I64d%I64d",&a,&p,&b)!=EOF) {if (b>=p) {printf("Orz,I can’t find D!\n");continue;}LL t=exbsgs(a,b,p);if (t!=-1) printf("%I64d\n",t);else printf("Orz,I can’t find D!\n");}}

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