C++博客-^^想你-文章分类-数论

ZOJ 3233 Lucky Number 容斥原理

Huicpc217 — Mon, 31 Aug 2009 17:28:00 GMT

#include<iostream>
using namespace std;

typedef long long ll;
ll a[20], b[1000], sum;

inline ll gcd( ll a, ll b ){
    ll t;
    while ( b ){
        t = a % b;  a = b;  b = t;
    }
    return a;
}

inline ll lcm( ll a, ll b ){
    return b / gcd( a, b ) * a ;
}

inline ll cel ( ll a, ll b, ll num ){
    return num / a - num / lcm( a, b );
}

void dfs( ll k, ll &n, ll &q, ll &num, ll id, ll now )
{
    if ( k == n || id == n ){
        return ;
    }
    ll i;
    for ( i = id ; i < n ; i ++ ){
        if ( k & 1 )
            sum -= cel( lcm( now, a[i] ), q, num );
        else
            sum += cel( lcm( now, a[i] ), q, num );
        dfs( k + 1, n, q, num, i + 1, lcm( now, a[i] ) );
    }
}

inline ll work( ll p[], ll n, ll q, ll num){
    sum = 0;
    dfs( 0, n, q, num, 0, 1 );
    return sum;
}

int main()
{
    ll i, j, k;
    ll n, m;
    ll st, ed;
    ll ans;
    while ( scanf("%lld%lld%lld%lld",&n,&m,&st,&ed) != EOF )
    {
        if ( ! ( n || m || st || ed ) )
            return 0;
        for ( i = 0 ; i < n ; i ++ )
            scanf("%lld",&a[i]);
        for ( i = 0 ; i < m ; i ++ )
            scanf("%lld",&b[i]);
        ll q = 1;
        for ( i = 0 ; i < m ; i ++ ){
            if ( q <= ed )
            q = lcm( b[i], q );
        }
        ans = work( a, n, q, st - 1  );
        cout<< ( work( a, n, q, ed ) - ans  ) <<endl;
    }
    return 0;
}

Huicpc217 2009-09-01 01:28 发表评论

10070 RSA

Huicpc217 — Wed, 26 Aug 2009 05:35:00 GMT

#include<iostream>
using namespace std;

__int64 mult( __int64 a, __int64 b, __int64 n)
{
    if( !b ) return 0;
    return ( mult( a, b >> 1 , n ) * 2 + a * ( b & 1 ) ) % n;
}

__int64 mod_pow( __int64 m,__int64 e,__int64 n)
{
    __int64 rel = 1;
    while ( e )
    {
        if ( e & 1 )
            rel = mult ( rel , m , n );
        m = mult( m, m, n );
        e >>= 1;
    }
    return rel;
}

__int64 ext_gcd( __int64 a,__int64 b,__int64 &x, __int64 &y )
{
    if ( ! b )
    {
        x = 1 ;
        y = 0 ;
        return a;
    }
    __int64 d = ext_gcd( b, a % b, x, y );
    __int64 t = x;
    x = y;
    y = t - a / b * y;
    return d;
}

__int64 gcd( __int64 a, __int64 b )
{
    __int64 t ;
    while ( b )
    {
        t = a ;
        a = b ;
        b = t % b ;
    }
    return a;
}

__int64 pollard( __int64 n )
{
    __int64 x = 1.0 * rand() / RAND_MAX * n;
    __int64 y = x, i = 1, k = 2 ;
    while ( true )
    {
        i ++ ;
        x = ( mult( x, x, n ) + 1 ) % n ;
        __int64 d ;
        if ( x > y )
            d = gcd( x - y , n );
        else
            d = gcd( y - x , n );
        if ( d > 1 && d < n )
            return d;
        if ( i == k )
        {
            y = x;
            k *= 2 ;
        }
    }
    return 1;
}

int main()
{
    __int64 c, e, n, m, d, r;
    srand(time(0));
    while ( scanf("%I64d%I64d%I64d",&c,&e,&n) != EOF )
    {
        __int64 p = pollard( n );
        __int64 fi_n = ( p - 1 ) * ( n / p - 1 );

        ext_gcd( e, fi_n, d, r);
        if( d < 0 ){
            d %= fi_n;  d += fi_n;
        }

        printf("%I64d\n",mod_pow( c, d, n ) );

    }
    return 0;
}

Huicpc217 2009-08-26 13:35 发表评论

HOJ 10007 Prime test miller_rabin + parllord

Huicpc217 — Wed, 26 Aug 2009 05:33:00 GMT

#include<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<time.h>
using namespace std;

typedef unsigned long long llu;

int p[] = { 2,3,5,7,11, 13,17,19,23,29 };
int k = 10;

// 不适合时限较低的题目。常数非常大！！！
llu mol_mult( llu a, llu b, llu n)
{
    if( !b ) return 0;
    return ( mol_mult( a, b >> 1 , n ) * 2 + a * ( b & 1 ) ) % n;
}
/*
// 代码量大，但常数低
llu mol_mult( llu a,llu b,const llu &mol)
{
    llu c = a >> 30 , d = b >> 30 , tmp , t;
    tmp = 1ll << 30;
    a %= tmp ; b %= tmp;

    llu ans = a * b;
    if( ans > mol )
        ans %= mol;

    tmp = c * b ;
    if( tmp > mol ) tmp %= mol;
    t = 30;
    while( t -- ){
        tmp <<= 1;
        if( tmp > mol ) tmp -= mol;
    }
    ans += tmp;
    if( ans > mol )
        ans -= mol;


    tmp = d * a ;
    if( tmp > mol ) tmp %= mol;
    t = 30;
    while( t -- ){
        tmp <<= 1;
        if( tmp > mol ) tmp -= mol;
    }
    ans += tmp;
    if( ans > mol )
        ans -= mol;

    tmp = d * c ;
    if( tmp > mol ) tmp %= mol;
    t = 60;
    while( t -- ){
        tmp <<= 1;
        if( tmp > mol ) tmp -= mol;
    }
    ans += tmp;

    ans %= mol;

    return ans;
}
*/
llu mol_exp( llu a,llu b,llu n)
{
    llu t = a % n , ans = 1;
    while( b )
    {
        if( b & 1 )
        {
            ans = mol_mult( ans, t, n);//1ll * ans * t % n;
        }
        t = mol_mult( t, t, n );//1ll * t * t % n;
        b >>= 1;
    }
    return ans;
}

int witness( llu a,llu n )
{
    llu t = 0, tmp = n - 1;
    while( tmp % 2 == 0 ){
        t ++ ;
        tmp >>= 1 ;
    }

    llu x0 = mol_exp( a, tmp, n ), x1, i;
    for ( i = 1 ; i <= t ; i ++ )
    {
        x1 = mol_mult( x0, x0 , n);//1ll * x0 * x0 % n;
            if( x1 == 1 && x0 != 1&& x0 != n - 1 )
                return 1;
        x0 = x1;
    }

    if( x1 != 1 )
        return 1;
    return 0;
}

inline llu my_rand(llu b,llu e )
{
    return (llu)((1. * rand() / ( 1. + RAND_MAX ) * (e - b + 1))) + b;
}

int miller_rabin( llu n,llu s )
{
    llu i, a;
    for ( i = 1 ; i <= s ; i ++ )
    {
        a = my_rand( 2, n - 1);
        if( witness ( a, n ) ) return 0;
    }
    return 1;
}

llu gcd(llu a,llu b){
    llu t ;
    while( b ){
        t = a % b; a = b; b = t;
    }
    return a;
}

llu pollard_rho( llu n, llu c )
{
    llu i = 1;
    llu x = my_rand( 1, n - 1 ), d;
    llu y = x;
    llu k = 2;
    while( 1 ){
        i ++;
        x = ( mol_mult( x , x, n ) + n - c ) % n;
        if( x == y )
            break;
        if ( x > y )
            d = gcd( x - y, n );
        else
            d = gcd( y - x, n );
        if( d != 1 && d != n )
            return d;
        if( i == k ){
            y = x;
            k <<= 1 ;
        }
    }
    return n;
}

llu minfactor;

void make(llu n)
{
    llu d, c;
    if( miller_rabin( n, 20 ) ){
        minfactor = min( minfactor , n );
        return ;
    }
    do{
        c = my_rand( 1, n - 1 );
        d = pollard_rho( n, c );
    }while( d >= n );
    make( d );
    make( n/d );
}

int main()
{
    llu n, i;
    srand( n );
    int test;
    cin>>test;
    while( test--)
    {
        cin>>n;
        if( n == 1 ){ puts("1"); continue; }
        for ( i = 0 ; i < k ; i ++ )
            if( n % p[i] == 0 ){
                if( n == p[i] )
                    puts("Prime");
                else
                    printf("%d\n",p[i]);
                break;
            }
        if( i < k ) continue;
        minfactor = n;
        if( miller_rabin( n, 20 ) )
            printf("Prime\n");
        else{
            make( n );
            printf("%llu\n",minfactor);
        }
    }
    return 0;
}

Huicpc217 2009-08-26 13:33 发表评论

高斯消元 Java 高精度版 HDU 2449 Gauss Elimination

Huicpc217 — Fri, 21 Aug 2009 18:36:00 GMT

import java.util.*;
import java.math.*;

class fraction{
    BigInteger a, b;
    public fraction(){
        a = new BigInteger("0");
        b = new BigInteger("1");
    }

    fraction( BigInteger a0, BigInteger b0){
        this.a = a0; this.b = b0;
    }
    void reduction(){
        BigInteger tmp = a.gcd( b );
        a = a.divide( tmp );
        b = b.divide( tmp );
        if( b.compareTo( BigInteger.ZERO ) == - 1 )
        {
            b = b.multiply( BigInteger.valueOf( -1 ));
            a = a.multiply( BigInteger.valueOf( -1 ));
        }
    }
    fraction add( fraction t ){
        fraction tmp = new fraction( a.multiply( t.b ).add( b.multiply( t.a )) , b.multiply(t.b) );
        tmp.reduction();
        return tmp;
    }
    fraction sub( fraction t ){
        fraction tmp = new fraction( a.multiply( t.b ).subtract( b.multiply( t.a )) , b.multiply(t.b) );
        tmp.reduction();
        return tmp;
    }
    fraction mult( fraction t){
        fraction tmp = new fraction( a.multiply( t.a ), b.multiply( t.b ));
        tmp.reduction();
        return tmp;
    }
    fraction div( fraction t){
        fraction tmp = new fraction( a.multiply( t.b ), b.multiply( t.a ));
        tmp.reduction();
        return tmp;
    }
    public void abs(){
        if( this.a.compareTo( BigInteger.ZERO ) == - 1 ){
            this.a = this.a.multiply( BigInteger.valueOf( -1 ));
        }
    }
    void out(){
        this.reduction();
        if( b.compareTo( BigInteger.ONE ) == 0 )
            System.out.println(a);
        else
            System.out.println(a+"/"+b);
    }

    boolean biger( fraction p ){
        fraction tmp = new fraction ( a, b );
        fraction t = new fraction(p.a,p.b);
        //t = p;
        tmp.reduction();
        if( tmp.a.compareTo( BigInteger.ZERO ) == - 1 ){
            tmp.a = tmp.a.multiply( BigInteger.valueOf( -1 ));
        }
        if( t.a.compareTo( BigInteger.ZERO ) == - 1 ){
            t.a = t.a.multiply( BigInteger.valueOf( -1 ));
        }
        tmp = tmp.sub( t );
        return tmp.a.compareTo( BigInteger.ZERO ) > -1;
    }

}

public class Main{

    public static void lup_solve( fraction x[],fraction y[], fraction L[][], fraction U[][],int pi[],fraction b[],int n)
    {
        int i, j;
        fraction z = new fraction( BigInteger.ZERO , BigInteger.ONE);
        fraction sum = z;//double sum;
        for ( i = 0 ; i < n ; i ++ ){
            sum = z; //sum = 0;
            for ( j = 0 ; j < i ; j ++ ){
                sum = sum.add( L[i][j].mult( y[j] ));//sum += L[i][j] * y[ j ];
            }
            y[i] = b[ pi[i] ].sub( sum );//y[i] = b[ pi[i] ] - sum;
        }
        for ( i = n - 1 ; i >= 0 ; i -- ){
            sum = z ; //sum = 0;
            for ( j = i + 1 ; j < n ; j ++ ){
                sum = sum.add( U[i][j].mult( x[j] ));//sum += U[i][j] * x[ j ];
            }
            x[i] = (y[i].sub( sum )).div( U[i][i] );//x[i] = (y[i] - sum)/U[i][i];
        }
    }

    public static int lup_decomposition( fraction a[][] , int n , int pi[] )
    {
        int i, j, k, k1 = 0 ;
        fraction p = new fraction(BigInteger.valueOf(0), BigInteger.ONE ), z = new fraction( BigInteger.valueOf(0), BigInteger.ONE );
        for ( i = 0 ; i < n ; i ++ )
            pi[i] = i;// 置换

        for ( k = 0 ; k < n ; k ++ ){
            p = z;
            for ( i = k ; i < n ; i ++ )
            {
                if(  a[i][k].biger( p ) )
                {
                    p = new fraction( a[i][k].a,a[i][k].b) ;
                    k1 = i;
                }
            }
            if( p.a.compareTo( BigInteger.ZERO ) == 0 ){
                return 0 ;// error
            }
            fraction tmp;

            int t = pi[ k ]; pi[ k ] = pi[ k1 ]; pi[k1] = t;
            for ( i = 0 ; i < n ; i ++ ){
                tmp = a[ k ][i];  a[ k ][i] = a[ k1 ][i];  a[k1][i] = tmp;
            }//swap( a[k][i], a[k1][i] );
            for ( i = k + 1 ; i < n ; i ++ )
            {
                a[i][k] = a[i][k].div( a[k][k] );
                for ( j = k + 1 ; j < n ; j ++ )
                    a[i][j] = a[i][j].sub( a[i][k].mult(a[k][j]));// - a[i][k] * a[k][j] ;
            }
        }
        return 1;
    }

    public static void check(fraction a[][], fraction x[], int n){
        int i, j;
        fraction sum, z = new fraction( BigInteger.ZERO , BigInteger.ONE);
        for ( i = 0 ;  i < n ; i++ ){
            sum = z;
            for ( j = 0 ;j < n ; j ++ )
            {
                sum = sum.add( a[i][j].mult( x[j] ));
            }
            sum.out();
        }
    }

    public static void main(String[] agrs){
        Scanner cin = new Scanner( System.in );
        int i, j;
        int n;
        while( cin.hasNextInt() )
        {
            //任何函数都要和一个class相连
            n = cin.nextInt();
            int pi[] = new int[n];
            fraction a[][] = new fraction[n][n];
            fraction aa[][] = new fraction[n][n];
            fraction B[] = new fraction[n];
            fraction x[] = new fraction[n];
            fraction y[] = new fraction[n];

            for ( i = 0 ; i < n ; i ++ )
            {
                for ( j = 0 ;j < n ; j ++ ){
                    a[i][j] = new fraction( BigInteger.valueOf(0),BigInteger.valueOf(1));
                    a[i][j].a = cin.nextBigInteger();
                    aa[i][j] = new fraction( BigInteger.valueOf(0),BigInteger.valueOf(1));
                    aa[i][j] = a[i][j];
                }
                B[i] = new fraction( BigInteger.valueOf(0),BigInteger.valueOf(1));
                B[i].a = cin.nextBigInteger();
                x[i] = new fraction( BigInteger.valueOf(0),BigInteger.valueOf(1));
                y[i] = new fraction( BigInteger.valueOf(0),BigInteger.valueOf(1));
            }
            if( 1 == lup_decomposition( a, n, pi) )
            {
                lup_solve( x, y, a, a, pi, B, n);
                for ( i = 0 ;i < n; i ++)
                    x[i].out();
                //check( aa, x, n);
            }
            else
            {
                System.out.println("No solution.");
            }
            System.out.println();
        }
    }
}

Huicpc217 2009-08-22 02:36 发表评论

FZU 1649 Prime number or not 63位素数判定

Huicpc217 — Wed, 19 Aug 2009 13:23:00 GMT

#include<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<time.h>
using namespace std;

typedef unsigned long long llu;

int p[] = { 2,3,5,7,11, 13,17,19,23,29 };
int k = 10;

llu mol_mult( llu a,llu b,const llu &mol)
{
    llu c = a >> 30 , d = b >> 30 , tmp , t;
    tmp = 1ll << 30;
    a %= tmp ; b %= tmp;

    llu ans = a * b;
    if( ans > mol )
        ans %= mol;

    tmp = c * b ;
    if( tmp > mol ) tmp %= mol;
    t = 30;
    while( t -- ){
        tmp <<= 1;
        if( tmp > mol ) tmp -= mol;
    }
    ans += tmp;
    if( ans > mol )
        ans -= mol;


    tmp = d * a ;
    if( tmp > mol ) tmp %= mol;
    t = 30;
    while( t -- ){
        tmp <<= 1;
        if( tmp > mol ) tmp -= mol;
    }
    ans += tmp;
    if( ans > mol )
        ans -= mol;

    tmp = d * c ;
    if( tmp > mol ) tmp %= mol;
    t = 60;
    while( t -- ){
        tmp <<= 1;
        if( tmp > mol ) tmp -= mol;
    }
    ans += tmp;

    ans %= mol;

    return ans;
}

llu mol_exp( llu a,llu b,llu n)
{
    llu t = a % n , ans = 1;
    while( b )
    {
        if( b & 1 )
        {
            ans = mol_mult( ans, t, n);//1ll * ans * t % n;
        }
        t = mol_mult( t, t, n );//1ll * t * t % n;
        b >>= 1;
    }
    return ans;
}

int witness( llu a,llu n )
{
    llu t = 0, tmp = n - 1;
    while( tmp % 2 == 0 ){
        t ++ ;
        tmp >>= 1 ;
    }

    llu x0 = mol_exp( a, tmp, n ), x1, i;
    for ( i = 1 ; i <= t ; i ++ )
    {
        x1 = mol_mult( x0, x0 , n);//1ll * x0 * x0 % n;
            if( x1 == 1 && x0 != 1&& x0 != n - 1 )
                return 1;
        x0 = x1;
    }

    if( x1 != 1 )
        return 1;
    return 0;
}

llu my_rand(llu b,llu e )
{
    return (llu)((1. * rand() / ( 1. + RAND_MAX ) * (e - b + 1))) + b;
}

int miller_rabin( llu n,llu s )
{
    llu i, a;
    for ( i = 1 ; i <= s ; i ++ )
    {
        a = my_rand( 2, n - 1);
        if( witness ( a, n ) ) return 0;
    }
    return 1;
}

int main()
{
    llu n, i;
    srand( n );
    while( scanf("%llu",&n) != EOF )
    {
        if( n == 1 ){ puts("It is not a prime number."); continue; }
        for ( i = 0 ; i < k ; i ++ )
            if( n % p[i] == 0 ){
                printf("%s\n", n == p[i] ? "It is a prime number.":"It is not a prime number." );
                break;
            }
        if( i < k ) continue;
        if( miller_rabin( n, 1 ) )
            printf("It is a prime number.\n");
        else
            printf("It is not a prime number.\n");
    }
    return 0;
}

Huicpc217 2009-08-19 21:23 发表评论

Hoj 11463 Color the necklace Polya

Huicpc217 — Wed, 19 Aug 2009 07:16:00 GMT

#include<iostream>
#include<vector>
using namespace std;

const int N = 100000;
const int mol = 20090531;

int a[N], p[N/10];
int p_fac[10], p_cnt[10], p_k;
int pf1[10], pc1[10], pk1;

int ans, m;

int prime2( int n)
{
    int i, j, k, t;
    for ( i = 2 , k = 0 ; i < n ; i ++ )
    {
        if( !a[i] )
            p[ k ++ ] = a[i] = i;
        for ( j = 0 ; j < k ; j ++ )
        {
            t = p[j] * i;
            if ( t >= n )
                break;
            a[ t ] = p[ j ];
            if( i % p[j] == 0 )
                break;
        }
    }
    return k;
}

// 要求 n > 1
void fac_break( int n, int p_fac[], int p_cnt[], int &p_k )
{
    if( n < N ){
        int t ;
        p_k = p_cnt[ 0 ] = 0;
        while( n != 1 )
        {
            t = p_fac[ p_k ] = a[n];
            while( n % t == 0 )
            {
                n /= t , p_cnt[ p_k ] ++ ;
            }
            p_k ++;
            p_cnt[ p_k ] = 0;
        }
    }
    else
    {
        p_k = p_cnt [ 0 ] = 0;
        for ( int i = 2 ; i * i <= n ; i ++ )
        {
            if( n % i == 0 )
            {
                p_fac[ p_k ] = i;
                while ( n % i == 0 )
                {
                    p_cnt [ p_k ] ++;
                    n /= i;
                }
                p_k ++;
                p_cnt[ p_k ] = 0;
            }
        }
        if( n != 1 ){
            p_fac[ p_k ] = n;
            p_cnt[ p_k ++ ] = 1;
        }
    }
}

int ruler( int n)
{
    fac_break( n , p_fac, p_cnt, p_k );
    for ( int i = 0 ; i < p_k ; i ++ ){
        n /= p_fac[ i ];
        n *= p_fac[ i ] - 1;
    }
    return n;
}

int mol_pow( int a, int b )
{
    int t = a, ans = 1;
    while( b ){
        if( b & 1 ){
            ans = 1ll * ans * t % mol;
        }
        t = 1ll * t * t % mol;
        b >>= 1;
    }
    return ans;
}

// 生成所有约数。已经调用了 fac_break();
void make_div( int k, int val, int p_fac[], int p_cnt[], const int &p_k, const int &n,const int &m ){
    if( k == p_k ){

        int t = ruler( n / val );
        ans = ( 1ll * t * mol_pow( m, val ) % mol + ans ) % mol;

        return ;
    }
    int t = 1;
    for ( int i = 0 ; i <= p_cnt[ k ] ; i ++ )
    {
        make_div( k + 1, val * t , p_fac, p_cnt, p_k, n, m );
        t *= p_fac[ k ];
    }
}

int ext_gcd( int a,int b,int &x,int &y)
{
    if( b == 0 )
    {
        x = 1, y = 0;
        return a;
    }
    else
    {
        int d = ext_gcd( b, a%b, x, y );
        int t = x;
        x = y;
        y = t - a/b*y;
        return d;
    }
}

int ni( int n )
{
    int x, y;
    ext_gcd( n, mol, x, y );
    x %= mol;
    if( x < 0 )
        x += mol;
    return x;
}

int main()
{
    prime2( N );
    int i, j, k, n;
    int test ;
    scanf("%d",&test);
    while( test -- )
    {
        scanf("%d",&n);
        m = n;
        ans = 0;
        fac_break( n, pf1, pc1, pk1 );
        make_div( 0, 1 , pf1, pc1, pk1, n, m );
        if( n % 2 ){
            ans = ( 1ll * n * mol_pow( m, n/2 + 1 ) % mol + ans ) % mol ;
        }
        else{
            ans = ( 1ll * n / 2 * mol_pow( m, n/2 + 1 ) % mol + 1ll * n / 2 * mol_pow( m, n / 2 ) % mol + ans ) % mol ;
        }
        //ni( n * 2 );
        ans = 1ll*ans*ni(n*2%mol)%mol;
        printf("%d\n",ans);
    }
    return 0;
}