POJ 3243 Clever Y

 

 

Time Limit: 5000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

Description

Little Y finds there is a very interesting formula in mathematics:

X^Y mod Z = K

Given X, Y, Z, we all know how to figure out K fast. However, given X, Z, K, could you figure out Y fast?

Input

Input data consists of no more than 20 test cases. For each test case, there would be only one line containing 3 integers X, Z, K (0 ≤ X, Z, K ≤ 10 ⁹).
Input file ends with 3 zeros separated by spaces.

Output

For each test case output one line. Write "No Solution" (without quotes) if you cannot find a feasible Y (0 ≤ Y < Z). Otherwise output the minimum Y you find.

Sample Input

5 58 33
2 4 3
0 0 0

Sample Output

9
No Solution

Source

POJ Monthly--2007.07.08, Guo, Huayang

 

依舊是離散對數。

和POJ2417基本相同：

http://www.cnblogs.com/SilverNebula/p/5668380.html

不一樣之處是這道題的X Z K沒有特殊限制，可能不是質數，傳統的作法應該是不斷轉化式子直到gcd(a',n')==1 ，而後再BSGS

代碼參照Orion_Rigel的題解http://blog.csdn.net/orion_rigel/article/details/51893151

 

然而套用了以前2417的代碼，發現能AC

也是神奇

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<cmath>
 6 #define LL long long
 7 using namespace std;
 8 const int mop=100007;
 9 const int mxn=100020;
10 int hash[mxn],hd[mxn],id[mxn],next[mxn],cnt;
11 
12 LL gcd(LL a,LL b){
13     return (b==0)?a:gcd(b,a%b);
14 }
15 void ins(LL x,LL y){
16     int k=x%mop;
17     hash[++cnt]=x;
18     id[cnt]=y;next[cnt]=hd[k];hd[k]=cnt;
19     return;
20 }
21 LL find(LL x){
22     int k=x%mop;
23     for(int i=hd[k];i;i=next[i]){
24         if(hash[i]==x)return id[i];
25     }
26     return -1;
27 }
28 LL BSGS(int a,int b,int n){
29     memset(hd,0,sizeof hd);
30     cnt=0;
31     if(!b)return 0;
32     int m=sqrt(n),i,j;
33     LL x=1,p=1;
34     for(i=0;i<m;i++,p=p*a%n)ins(p*b%n,i);
35     for(LL i=m;;i+=m){
36         x=x*p%n;
37         if((j=find(x))!=-1)return i-j;
38         if(i>n)break;
39     }
40     return -1;
41 }
42 int main(){
43     int x,z,k;
44     while(scanf("%d%d%d",&x,&z,&k)!=EOF){
45         if(!x && !z && !k)break;
46         int ans=BSGS(x,k,z);
47         if(ans==-1)printf("No Solution\n");
48         else printf("%d\n",ans);
49         
50     }
51     return 0;
52 }