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HDU 1452 Happy 2004(因子�?

Vontroy — Sat, 02 Oct 2010 14:08:00 GMT

/*************************************************************
计算 2004^X的因子和 s(2004^X)   mod M, M=29

  s(2004^X)%29
  因子�?nbsp;s是积性函�?�?nbsp;�Q�gcd(a,b)=1==> s(a*b)= s(a)*s(b)

  2004^X=4^X * 3^X *167^X
  s(2004^X)=  s(2^(2X))* s(3^X) * s(167^X)

  如果 p是素�?nbsp;==> s(p^X)=1+p+p^2++p^X = (p^(X+1)-1) /(p-1)

  s(2004^X)=�Q?^(2X+1)-1�Q? (3^(X+1)-1)/2  *(167^(X+1)-1)/166

  167%29=22

  s(2004^X)=�Q?^(2X+1)-1�Q? (3^(X+1)-1)/2  *(22^(X+1)-1)/21

  (a*b)/c %M= a%M* b%M  * inv(c)

  c*inv(c)=1 %M    模�ؓ(f��)1的所有数  inv�Q�c�Q��ؓ(f��)最��可以被c整除�?br />
    inv(2)=15,  inv(21)=18    2*15=1 mod 29, 18*21=1 mod 29

  s(2004^X)=�Q?^(2X+1)-1�Q? (3^(X+1)-1)/2  *(22^(X+1)-1)/21
           =�Q?^(2X+1)-1�Q? (3^(X+1)-1)*15 *(22^(X+1)-1)*18
***************************************************************/

//hdu 1452
#include <iostream>
#include <cstdio>
#include <cmath>

using namespace std;

int _pow( int a, int n )
{
    int b = 1;
    while( n > 1 )
        if( n % 2 == 0 )
        {
            a = ( a * a ) % 29;
            n /= 2;
        }
        else
        {
            b = b * a % 29;
            n--;
        }
        return a * b % 29;
}
int main()
{
    int X;
    int a, b, c;
    while( cin >> X, X )
    {
        a = _pow( 2, 2 * X + 1 );
        b = _pow( 3, ( X + 1 ) );
        c = _pow( 22, ( X + 1 ) );

        cout << ( a - 1 ) * ( b - 1 ) * 15 * ( c - 1 ) * 18 % 29 << endl;
    }
    return 0;
}

Vontroy 2010-10-02 22:08 发表评论

POJ 1006 Biorhythms 中国剩余定理

Vontroy — Sat, 02 Oct 2010 12:18:00 GMT

/***************************************
比较�l�典的中国剩余定理~~

�?3×28�?3除余1�Q�用33×28×6=5544

�?3×33�?8除余1�Q�用23×33×19=14421

�?3×28�?3除余1�Q�用23×28×2=1288

�Q?544×p+14421×e+1288×i�Q?�Q?3×28×33�Q?n+d

n=�Q?544×p+14421×e+1288×i-d�Q?�Q?3×28×33�Q?br />****************************************/

#include <iostream>
#include <cstdio>

using std :: cin;
using std :: cout;
using std :: endl;

int get__( int a, int b, int c )
{
    int ans;
    int cnt = 1;
    while( ( a * b * cnt ) % c != 1 )
    {
        cnt++;
    }
    return a * b * cnt;
}

int main()
{
    int _p = get__( 28, 33, 23 );
    int _e = get__( 23, 33, 28 );
    int _i = get__( 23, 28, 33 );

    int p, e, i, d;
    int cas = 1;
    while( cin >> p >> e >> i >> d )
    {
        if( p == e && e == i && i == d && d == -1 ) break;
        int ans = ( _p * p + _e * e + _i * i - d ) % ( 23 * 28 * 33 );
        if( ans <= 0 )
            printf("Case %d: the next triple peak occurs in %d days.\n", cas++, 21252 - ans - 2 * d);
        else
            printf("Case %d: the next triple peak occurs in %d days.\n", cas++, ans);
    }
    return 0;
}

Vontroy 2010-10-02 20:18 发表评论

HDU 1788 Chinese remainder theorem again

Vontroy — Sat, 02 Oct 2010 06:58:00 GMT

/**********************************
N % MI = MI - a
因�ؓ(f��) a < MI
原式�{��h(hu��n)�?nbsp;(N + a) % MI = 0
所以此题�ؓ(f��)�?nbsp;M0 �?nbsp;MI 的最��公倍数

(注意�_�ֺ�问题,用__int64)

***********************************/

#include <iostream>
#include <cstdio>

using namespace std;

__int64 gcd( __int64 a, __int64 b )
{
    if( b == 0 ) return a;
    return gcd( b, a % b );
}

__int64 lcm( __int64 a, __int64 b )
{
    return a * b / gcd( a, b );
}

int main()
{
    int n, k;
    int tmp;
    __int64 ans;
    while( cin >> n >> k, n || k )
    {
        ans = 1;
        for( int i = 0; i < n; i++ )
        {
            cin >> tmp;
            ans = lcm( ans, tmp );
        }
        cout << ans - k << endl;
    }
    return 0;
}

Vontroy 2010-10-02 14:58 发表评论

Vontroy — Sat, 02 Oct 2010 06:28:00 GMT

摘要: //数论模板#include#includeusing namespace std;//辗�{盔R��法求最大公�U�数l(f��)ong gcd(long a, long b){ if(b==0) ... 阅读全文

Vontroy 2010-10-02 14:28 发表评论

HDU 1018 Big Number

Vontroy — Sat, 02 Oct 2010 06:22:00 GMT

斯特灵公�?/strong>是一条用来取n阶乘�q�似�?/font>�?font color="#000000">数学公式。一般来��_(d��)��当n很大的时候，n阶乘的计��量十分大，所以斯特灵公式十分好用�Q�而且�Q�即使在

n很小的时候，斯特灵公式的取值已�l�十分准��?/p>
公式为：(x��)

�q�就是说�Q�对于��够大的整�?em>n�Q�这两个��C��似倹{��更加精��地�Q?img border="0" alt="" src="http://m.shnenglu.com/images/cppblog_com/vontroy/2.png" width="194" height="65" />

或者：(x��)

Big Number

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8759    Accepted Submission(s): 3879

Problem Description

In many applications very large integers numbers are required. Some of these applications are using keys for secure transmission of data, encryption, etc. In this problem you are given a number, you have to determine the number of digits in the factorial of the number.

Input

Input consists of several lines of integer numbers. The first line contains an integer n, which is the number of cases to be tested, followed by n lines, one integer 1 ≤ n ≤ 10⁷ on each line.

Output

The output contains the number of digits in the factorial of the integers appearing in the input.

Sample Input

2 10 20

Sample Output

7 19

//log10(n!)=(0.5*log(2*PI*n)+n*log(n)-n)/log(10)

#include <iostream>
#include <cstdio>
#include <cmath>

const double PI = 3.1415926;

int main()
{
    int n;
    int tmp;
    while( ~scanf("%d", &n ) )
    {
        for( int i = 0; i < n; i++ )
        {
            scanf("%d", &tmp);
            double cnt = 1;
            cnt += (0.5 * log( 2 * PI * tmp ) + tmp * log( tmp ) - tmp ) / log(10);
            printf("%d\n", (int)(cnt));
        }
    }
    return 0;
}

Vontroy 2010-10-02 14:22 发表评论

Vontroy — Sat, 02 Oct 2010 06:20:00 GMT

求两个或N个数的最大公�U�数(gcd)和最��公倍数(lcm)的较优算�?/span>

//两个数的最大公�U�数--�Ƨ几里得��法

int gcd(int a, int b)

{

     if (a < b)

          swap(a, b);

     if (b == 0)

           return a;

      else

            return gcd(b, a%b);

}

//n个数的最大公�U�数��法

//说明:

//把n个数保存��Z��个数�l?br>
//参数为数�l�的指针和数�l�的大小(需要计��的数的个数)

//然后先求出gcd(a[0],a[1]), 然后��所求的gcd与数�l�的下一个元素作为gcd的参数��l�求gcd

//�q�样��׃�生一个递归的求ngcd的算�?/span>

int ngcd(int *a, int n)

{

    if (n == 1)  return *a;

    return gcd(a[n-1], ngcd(a, n-1));

}

//两个数的最��公倍数(lcm)��法

//lcm(a, b) = a*b/gcd(a, b)

int lcm(int a, int b)

{

        return a*b/gcd(a, b);

}

//n个数的最��公倍数��法

//��法�q�程和n个数的最大公�U�数求法�c�M��

//求出头两个的最��公倍数,再将�ƺ和大三个数求最��公倍数直到数组末尾

//�q�样产生一个递归的求nlcm的算�?/span>

int nlcm(int *a, int n)

{

      if (n == 1)

            return *a;

      else

            return lcm(a[n-1], nlcm(a, n-1));

}

Vontroy 2010-10-02 14:20 发表评论