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  1. 快速幂
  2. Exgcd

    xy \equiv z \pmod p

    yx+pb = z

    可以用Exgcd求解

  3. bsgs

    求解关于x的方程:

    y^x \equiv z \pmod p

    其中\gcd(y,p)=1

    解法:

    x=am-b

    那么y^{am} \equiv z \cdot y^b \pmod p

    右边:

    枚举b的取值[0,m-1],计算右边的值,存入哈希表

    左边:

    枚举可能的a,也就是[1,m],若左边的值在哈希表中出现过,那么当前a合法

    不难证明当m=\sqrt p的时候复杂度最优,\Theta(\sqrt p)

#include<bits/stdc++.h>
namespace ZDY{
    #pragma GCC optimize(3)
    #define il __inline__ __attribute__ ((always_inline))
    #define register
    #define ll long long
    #define ull unsigned long long
    #define db double
    #define sht short
    #define MB template <class T>il
    #define Fur(i,x,y) for(int i(x);i<=y;++i)
    #define Fdr(i,x,y) for(int i(x);i>=y;--i)
    #define fl(i,x) for(int i(head[x]),to;to=e[i].to,i;i=e[i].nxt)
    #define clr(x,y) memset(x,y,sizeof(x))
    #define cpy(x,y) memcpy(x,y,sizeof(x))
    #define fin(s) freopen(s".in","r",stdin)
    #define fout(s) freopen(s".out","w",stdout)
    #define fcin ios::sync_with_stdio(false)
    #define l2(n) ((int)(log2(n)))
    #define inf 2122219134
    MB T ABS(T x){return x>0?x:-x;}
    MB T MAX(T x,T y){return x>y?x:y;}
    MB T MIN(T x,T y){return x<y?x:y;}
    MB T GCD(T x,T y){return y?GCD(y,x%y):x;}
    MB void SWAP(T &x,T &y){T t=x;x=y;y=t;}
}using namespace ZDY;using namespace std;
namespace IO{const int str=1<<20;static char in_buf[str],*in_s,*in_t;bool __=0;il char gc(){return (in_s==in_t)&&(in_t=(in_s=in_buf)+fread(in_buf,1,str,stdin)),in_s==in_t?__=1,EOF:*in_s++;}il void in(string &ch){ch.clear();if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF){__=1;return;}ch+=c;while((c=gc())!=EOF&&!isspace(c))ch+=c;if(c==EOF)__=1;}il void in(char &ch){if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)__=1;else ch=c;}il void in(char *ch){*ch='\0';if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF){__=1;return;}*ch=c;ch++;while((c=gc())!=EOF&&!isspace(c))*ch=c,ch++;if(c==EOF)__=1;*ch='\0';}template<typename T>il void in(T &x){if(__)return;char c=gc();bool f=0;while(c!=EOF&&(c<'0'||c>'9'))f^=(c=='-'),c=gc();if(c==EOF){__=1;return;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=x*10+c-48,c=gc();if(c==EOF)__=1;if(f)x=-x;}template<typename T,typename ... arr>il void in(T &x,arr & ... y){in(x),in(y...);}const char ln='\n';static char out_buf[str],*out_s=out_buf,*out_t=out_buf+str;il void flush(){fwrite(out_buf,1,out_s-out_buf,stdout);out_s=out_buf;}il void pt(char c){(out_s==out_t)?(fwrite(out_s=out_buf,1,str,stdout),*out_s++=c):(*out_s++=c);}il void out(const char* s){while(*s)pt(*s++);}il void out(char* s){while(*s)pt(*s++);}il void out(char c){pt(c);}il void out(string s){for(int i=0;s[i];i++)pt(s[i]);}template<typename T>il void out(T x){if(!x){pt('0');return;}if(x<0)pt('-'),x=-x;char a[50],t=0;while(x)a[t++]=x%10,x/= 10;while(t--)pt(a[t]+'0');}template<typename T,typename ... arr>il void out(T x,arr & ... y){out(x),out(y...);}}using namespace IO;
int T,typ,p;
il int pw(int x,int b){
    int ans=1;
    while(b){
        if(b&1)ans=1ll*ans*x%p;
        x=1ll*x*x%p;b>>=1;
    }
    return ans;
}
struct hash{
    #define N 37273
    int P=N,head[N],cnt;
    struct edge{int to,nxt,w;}e[N];
    il void clear(){cnt=0;clr(head,0);}
    il void add(int x,int y,int w){e[++cnt].to=y;e[cnt].nxt=head[x];head[x]=cnt;e[cnt].w=w;}
    il void ins(int x,int v){add(x%P,x,v);}
    int operator[](int x){
        for(int i=head[x%P];i;i=e[i].nxt)
            if(e[i].to==x)return e[i].w;
        return -1;
    }
}S;
int main(){
    in(T,typ);
    while(T--){
        int a,b,t;
        in(a,b,p);
        if(typ==1)out(pw(a%p,b),ln);
        if(typ==2){//xa equiv b
            a%=p;b%=p;
            if(!a&&b)out("Orz, I cannot find x!\n");
            else out(1ll*pw(a,p-2)*b%p,ln);
        }
        if(typ==3){//a^x equiv b;
            a%=p;
            if(!a){
                if(!b)out("1\n");
                else out("Orz, I cannot find x!\n");
                continue;
            }
            int m=sqrt(p)+1,x=b%p,t=pw(a,m);
            S.clear();
            S.ins(x,0);
            Fur(i,1,m)x=1ll*x*a%p,S.ins(x,i);
            x=1;
            bool ff=1;
            Fur(i,1,m){
                x=1ll*x*t%p;
                int k=S[x];
                if(~k){
                    ff=0;
                    out(((i*m-k)%p+p)%p,ln);
                    break;
                }
            }
            if(ff)out("Orz, I cannot find x!\n");
        }
    }
    flush();
}
LG 2485 [SDOI2011]计算器
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