Baby Step Giant Step

来源：互联网发布：淘宝直播人数真的吗编辑：程序博客网时间：2024/05/17 08:32

Baby Step Giant Step 模板题

POJ 2417 基础 Baby Step Giant Step http://poj.org/problem?id=2417

HDU 2815 扩展 Baby Step Giant Step http://acm.hdu.edu.cn/showproblem.php?pid=2815

POJ 3243 扩展 Baby Step Giant Step http://poj.org/problem?id=3243

扩展 Baby Step Giant Step 详细请看 http://hi.baidu.com/aekdycoin/item/236937318413c680c2cf29d4 - By AekdyCoin

POJ 2417 http://poj.org/problem?id=2417

#include<stdio.h>#include<algorithm>#include<string.h>#include<math.h>#include<map>using namespace std;typedef long long ll;void exgcd(int a,int b,int &x,int &y){    if(b){exgcd(b,a%b,y,x);y-=(a/b)*x;}    else{x=1;y=0;}}int gcd(int a,int b){return b?gcd(b,a%b):a;}#define mod 131071int Hash1[mod];int Hash2[mod];int getHash(int n){    int tem=n%mod;    while(Hash1[tem]!=-1){        if(Hash1[tem]!=n){            tem++;            if(tem>=mod)                tem-=mod;        }        else break;    }    return tem;}inline int BSGS(int A,int B,int C){ // A^x=B % C    ll tem=1%C;    for(int i=0;i<=100;i++)        if(tem==B) return i;        else tem=tem*A%C;        ll D=1%C;    int d=0;    while((tem=gcd(A,C))!=1){        if(B%tem!=0) return -1;        d++;        C/=tem;        B/=tem;        D=(D*A/tem)%C;    }        memset(Hash1,-1,sizeof(Hash1));    memset(Hash2,-1,sizeof(Hash2));    int m=(int)ceil(sqrt((double)C));        tem=1%C;    ll K;    for(int i=0;i<=m;i++){        if(i==m) K=tem;        int h=getHash((int)tem);        if(Hash1[h]==-1) Hash1[h]=(int)tem,Hash2[h]=i;        tem=(tem*A)%C;    }        for(int i=0;i<=m;i++){        int u,v;        exgcd((int)D,C,u,v);        u=(((ll)u*B)%C+C)%C;        int h=getHash(u);        if(u>=0&&Hash1[h]!=-1)            return i*m+Hash2[h]+d;        D=D*K%C;    }    return -1;}int main(){    int A,B,C;    while(scanf("%d%d%d",&C,&A,&B)!=EOF){// A^x=B % C        int ans;        if((ans=BSGS(A,B,C))!=-1) printf("%d\n",ans);        else printf("no solution\n");    }}

HDU 2815 http://acm.hdu.edu.cn/showproblem.php?pid=2815

Hash版：

#include<stdio.h>#include<algorithm>#include<string.h>#include<math.h>#include<map>using namespace std;typedef long long ll;void exgcd(int a,int b,int &x,int &y){    if(b){exgcd(b,a%b,y,x);y-=(a/b)*x;}    else{x=1;y=0;}}int gcd(int a,int b){return b?gcd(b,a%b):a;}#define mod 131071int Hash1[mod];int Hash2[mod];int getHash(int n){    int tem=n%mod;    while(Hash1[tem]!=-1){        if(Hash1[tem]!=n){            tem++;            if(tem>=mod)                tem-=mod;        }        else break;    }    return tem;}inline int BSGS(int A,int B,int C){ // A^x=B % C    if(B>=C) return -1;    ll tem=1%C;    for(int i=0;i<=100;i++)        if(tem==B) return i;        else tem=tem*A%C;        ll D=1%C;    int d=0;    while((tem=gcd(A,C))!=1){        if(B%tem!=0) return -1;        d++;        C/=tem;        B/=tem;        D=(D*A/tem)%C;    }        memset(Hash1,-1,sizeof(Hash1));    memset(Hash2,-1,sizeof(Hash2));    int m=(int)ceil(sqrt((double)C));        tem=1%C;    ll K;    for(int i=0;i<=m;i++){        if(i==m) K=tem;        int h=getHash((int)tem);        if(Hash1[h]==-1) Hash1[h]=(int)tem,Hash2[h]=i;        tem=(tem*A)%C;    }        for(int i=0;i<=m;i++){        int u,v;        exgcd((int)D,C,u,v);        u=(((ll)u*B)%C+C)%C;        int h=getHash(u);        if(u>=0&&Hash1[h]!=-1)            return i*m+Hash2[h]+d;        D=D*K%C;    }    return -1;}int main(){    int A,B,C;    while(scanf("%d%d%d",&A,&C,&B)!=EOF){// A^x=B % C        int ans;        if((ans=BSGS(A,B,C))!=-1) printf("%d\n",ans);        else printf("Orz,I can’t find D!\n");    }}

map版：

#include<stdio.h>#include<algorithm>#include<string.h>#include<math.h>#include<map>using namespace std;typedef long long ll;void exgcd(int a,int b,int &x,int &y){    if(b){exgcd(b,a%b,y,x);y-=(a/b)*x;}    else{x=1;y=0;}}int gcd(int a,int b){return b?gcd(b,a%b):a;}map<int,int> Hash;inline int BSGS(int A,int B,int C){ // A^x=B % C    if(B>=C)return -1;    ll tem=1%C;    for(int i=0;i<=100;i++)         //处理出小的        if(tem==B) return i;        else tem=tem*A%C;        ll D=1%C;    int d=0;    while((tem=gcd(A,C))!=1){       //降次        if(B%tem!=0) return -1;        d++;        C/=tem;        B/=tem;        D=(D*A/tem)%C;    }        Hash.clear();    int m=(int)ceil(sqrt((double)C));        tem=1%C;    ll K;        for(int i=0;i<=m;i++){        if(i==m) K=tem;        if(Hash.find((int)tem)==Hash.end()) Hash[(int)tem]=i;        tem=(tem*A)%C;    }        for(int i=0;i<=m;i++){        int u,v;        exgcd((int)D,C,u,v);        u=(((ll)u*B)%C+C)%C;        if(u>=0&&Hash.find(u)!=Hash.end())            return i*m+Hash[u]+d;        D=D*K%C;    }    return -1;}int main(){    int A,B,C;    while(scanf("%d%d%d",&A,&C,&B)!=EOF){// A^x=B % C        int ans;        if((ans=BSGS(A,B,C))!=-1) printf("%d\n",ans);        else printf("Orz,I can’t find D!\n");    }}

0 0