标签：blank prim air expected some poj rar UI ase
BL
== N (mod P)
Input
Output
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
B(P-1)
== 1 (mod P)
B(-m)
== B(P-1-m)
(mod P) .
这道题是裸的BSGS，具体内容可以看hzw的博客—传送门
1 #include<algorithm> 2 #include<map> 3 #include<cstdio> 4 #include<cstring> 5 #include<cmath> 6 #define ll long long 7 using namespace std; 8 ll p,b,n,s,x,y,m,k; 9 int exgcd(ll a,ll b){ 10 if (!b){ 11 x=1; y=0; 12 return a; 13 } 14 int d=exgcd(b,a%b); 15 ll t=x; x=y; y=t-(a/b)*y; 16 return d; 17 } 18 map<int,int> h; 19 int main(){ 20 while (~scanf("%lld%lld%lld",&p,&b,&n)){ 21 h.clear(); 22 ll t=(ll)sqrt(p); 23 s=1; h[1]=t; 24 for (int i=1;i<=t-1;i++){ 25 s=s*b%p; 26 if (!h[s]) h[s]=i; 27 } 28 s=s*b%p; 29 ll l=1e10,ans=n; 30 exgcd(s,p); 31 x=(x+p)%p; 32 for (int i=0;i<=t;i++){ 33 if (h[ans]){ 34 if (h[ans]==t) h[ans]=0; 35 l=i*t+h[ans]; 36 break; 37 } 38 ans=ans*x%p; 39 } 40 if (l!=1e10) printf("%lld\n",l); 41 else puts("no solution"); 42 } 43 return 0; 44 }
POJ-2417-Discrete Logging（BSGS）
标签：blank prim air expected some poj rar UI ase
原文地址：http://www.cnblogs.com/zhuchenrui/p/7632822.html
