| Time Limit: 5000MS | Memory Limit: 65536KB | 64bit IO Format: %I64d & %I64u |
Description
Given a prime P, 2 <= P < 2 31, an integer B, 2 <= B < P, and an integer N, 1 <= N < P, compute the discrete logarithm of N, base B, modulo P. That is, find an integer L such that
B
L
== N (mod P)
Input
Read several lines of input, each containing P,B,N separated by a space.
Output
For each line print the logarithm on a separate line. If there are several, print the smallest; if there is none, print "no solution".
Sample Input
5 2 1
5 2 2
5 2 3
5 2 4
5 3 1
5 3 2
5 3 3
5 3 4
5 4 1
5 4 2
5 4 3
5 4 4
12345701 2 1111111
1111111121 65537 1111111111
Sample Output
0
1
3
2
0
3
1
2
0
no solution
no solution
1
9584351
462803587
Hint
The solution to this problem requires a well known result in number theory that is probably expected of you for Putnam but not ACM competitions. It is Fermat's theorem that states
for any prime P and some other (fairly rare) numbers known as base-B pseudoprimes. A rarer subset of the base-B pseudoprimes, known as Carmichael numbers, are pseudoprimes for every base between 2 and P-1. A corollary to Fermat's theorem is that for any m
B
(P-1)
== 1 (mod P)
for any prime P and some other (fairly rare) numbers known as base-B pseudoprimes. A rarer subset of the base-B pseudoprimes, known as Carmichael numbers, are pseudoprimes for every base between 2 and P-1. A corollary to Fermat's theorem is that for any m
B
(-m)
== B
(P-1-m)
(mod P) .
Source
式子是这个：BL == N (mod P) 求最小的L
把L分解：L=i*m+j 其中m=sqrt(P)向上取整
原式化为 B^(i*m)*B^j==N(MOD P)
再化为B^j===N*B^(-i*m)(MOD P)
先预处理出B^1 B^2 ... B^(m-1)
之后求出B^-m,枚举i，如果存在B^(-i*m)存在于HASH表中，说明存在解L=i*m+j
说了这么多，其实我也不怎么懂2333
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隔了一段时间回来，发现代码看不懂了，这和上面写的不一样啊？
研究了一下，发现确实不一样
按照上面写的解释，算N*B^(-i*m)很明显需要用到乘法逆元，这就比较麻烦。如何规避这个问题？
把L分解成L=i*m-j （反正m和j都是待求的，就当m比上面的分解法多了1吧）
↑这样原式化为B^(i*m) / B^j == N (MOD P)
B^(i*m) == N * B^j (MOD P)
这样就不需要求逆元咯。下面的代码就是这种写法
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代码基本靠抄
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<cmath> 6 using namespace std; 7 const int mop=100007; 8 int hash[mop]; 9 int head[mop],nxt[mop],id[mop],cnt; 10 void add(int x,int y){//hash 11 int m=x%mop; 12 hash[++cnt]=x;nxt[cnt]=head[m];id[cnt]=y;head[m]=cnt; 13 return; 14 } 15 int find(int x){ 16 int m=x%mop; 17 int i,j; 18 for(i=head[m];i!=-1;i=nxt[i]){ 19 if(hash[i]==x)return id[i]; 20 } 21 return -1; 22 } 23 int BSGS(int a,int b,int n){//a^x mod n =b 24 memset(head,-1,sizeof head); 25 cnt=0; 26 if(b==1)return 0; 27 int m=sqrt((double)n),j; 28 long long x=1,p=1; 29 for(int i=0;i<m;i++,p=p*a%n)add(p*b%n,i);//预处理b^i 30 for(long long i=m; ;i+=m){ 31 x=x*p%n; 32 if((j=find(x))!=-1)return i-j; 33 if(i>n)break; 34 } 35 return -1; 36 } 37 int main(){ 38 int p,b,n; 39 while(scanf("%d%d%d",&p,&b,&n)!=EOF){ 40 int ans=BSGS(b,n,p); 41 if(ans==-1)printf("no solution ");//无解 42 else printf("%d ",ans); 43 } 44 return 0; 45 }