BSGS算法是meet in the middle思想的一种应用,参考Yveh的博客我学会了BSGS的模版和hash表模板,,,
现在才会hash是不是太弱了,,,
#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct node{
static const int mo=100007;
int a[100010],v[100010];
node() {memset(a,-1,sizeof(a));}
int find(int val){
int pos=(val%mo+mo)%mo;
while ((a[pos]!=val)&&(a[pos]!=-1)) pos=(pos+1)%mo;
return pos;
}
void insert(int val,int x){
int pos=find(val);
if ((a[pos]==-1)||(a[pos]==val)){
a[pos]=val;
v[pos]=x;
}
}
int get(int val){
int pos=find(val);
return a[pos]==val?v[pos]:-1;
}
void clear() {memset(a,-1,sizeof(a));}
}hash;
int T,L;
inline int work1(int a,int b,int n){
long long t=a,ans=1;
while (b){
if (b%2==1) ans=(ans*t)%n;
b=b/2;
t=(t*t)%n;
}
return (int)ans;
}
inline int exgcd(int a,int b,int &x,int &y){
if (b==0){
x=1; y=0; return a;
}else{
int r=exgcd(b,a%b,x,y);
int t=y;
y=x-a/b*y;
x=t;
return r;
}
}
inline void work2(int a,int b,int n){
int x,y;
int d=exgcd(a,n,x,y);
if (b%d) puts("Orz, I cannot find x!\n");
else printf("%d\n",(int)((((long long)x*b/d%n)+n)%n));
}
inline void work3(int a,int b,int n){
if ((a%n==0)&&(b!=0)) {puts("Orz, I cannot find x!\n"); return;}
hash.clear();
int m=ceil(sqrt(n));
long long t=b%n;
for(int i=0;i<=m;++i){
hash.insert((int)t,i);
t=(t*a)%n;
}
int s=work1(a,m,n); t=s;
for(int i=1;i<=m;++i){
int v=hash.get((int)t);
if (v!=-1) {printf("%d\n",i*m-v); return;}
t=(t*s)%n;
}
puts("Orz, I cannot find x!\n");
}
int main(){
scanf("%d %d\n",&T,&L);
int y,z,p,a,b;
switch (L){
case 1:
while (T--){
scanf("%d %d %d\n",&y,&z,&p);
printf("%d\n",work1(y,z,p));
}
break;
case 2:
while (T--){
scanf("%d %d %d\n",&y,&z,&p);
work2(y,z,p);
}
break;
case 3:
while (T--){
scanf("%d %d %d\n",&y,&z,&p);
work3(y,z,p);
}
break;
}
return 0;
}
这样就可以啦