色综合久久天天综线观看,人妻无码久久精品,一级做a爰片久久毛片免费陪

luis — Tue, 10 Apr 2012 23:16:00 GMT

假如p是质敎ͼ��?a,p)=1�Q�那�?a^(p-1) ≡1�Q�mod p�Q?
假如p是质敎ͼ�且a,p互质�Q�那�?a�?p-1)�ơ方除以p的余数恒�{�于1
历史

　　与费马小定理相关的有一�?/span>中国猜想�Q�这个猜��x��中国数学家提出来的，其内容�ؓ(f��)�Q�当且仅�?^(p-1�Q?#8801;1(mod p�Q�，p是一个质数�?/span>

　　假如p是一个质数的话，�?^(p-1�Q?#8801;1(mod p�Q�成立（�q�是贚w��定理的一个特�D�情况）是对的。但反过来，假如2^(p-1�Q?#8801;1(mod p�Q�成立那么p是一个质数是不成立的�Q�比�?41�W�合上述条�g但不是一个质敎ͼ�。因此整个来说这个猜��x��错误的。一般认��Z��国数学家在费马前2000�q�的时候就已经认识中国猜测了，但也有�h认�ؓ(f��)实际上中国猜��是1872�q�提出的�Q�认为它早就��Z�h所知是��Z��一个误解�?/span>

�~�辑本段 证明

　　一、准备知识：(x��)

　　引理1�Q?a target="_blank" style="color: #136ec2; ">剩余�p?/a>定理2

　　若a,b,c��Z�Q�?个整敎ͼ�m为正整数�Q�且�Q�m,c)=1�Q�则当ac≡bc(mod m�Q�时�Q�有a≡b(mod m)

　　证明�Q�ac≡bc(mod m�Q�可得ac–bc≡0(mod m�Q�可得（a-b)c≡0(mod m�Q�因为（m,c)=1即m,c互质�Q�c可以�U�去�Q�a–b≡0(mod m�Q�可得a≡b(mod m)

　　引理2�Q�剩余系定理5

　　若m为整��C��m>1,a[1],a[2],a[3],a[4]�Q?#8230;a[m]为m个整敎ͼ�若在�q�m个数中�Q�?个整数对m不同余，则这m个整数对m构成完全剩余�p�R�?/span>

　　证明�Q�构造m的完全剩余系�Q?,1,2�Q?#8230;m-1�Q�，所有的整数必然�q�些整数中的1个对模m同余。取r[1]=0,r[2]=1,r[3]=2,r[4]=3�Q?#8230;r=i-1,1

　　引理3�Q�剩余系定理7

　　设m是一个整敎ͼ�且m>1�Q�b是一个整��C��Q�m,b)=1。如果a1,a2,a3,a4�Q?#8230;am是模m的一个完全剩余系�Q�则ba[1],ba[2],ba[3],ba[4]�Q?#8230;ba[m]也构成模m的一个完全剩余系�?/span>

　　证明�Q�若存在2个整数ba和ba[j]同余即ba≡ba[j](mod m�Q�，�Ҏ(gu��)��引理1则有a≡a[j](mod m�Q�。根据完全剩余系的定义和引理4�Q�完全剩余系中�Q�?个数之间不同余，易证明）可知�q�是不可能的�Q�因此不存在2个整数ba和ba[j]同余。由引理5可知ba[1],ba[2],ba[3],ba[4]�Q?#8230;ba[m]构成模m的一个完全剩余系�?/span>

　　引理4�Q?a target="_blank" style="color: #136ec2; ">同余定理6

　　如果a,b,c,d是四个整敎ͼ�且a≡b(mod m),c≡d(mod m�Q�，则有ac≡bd(mod m)

　　证明�Q�由题设得ac≡bc(mod m),bc≡bd(mod m�Q�，由模�q�算的传递性可得ac≡bd(mod m)

　　二、证明过�E�：(x��)

　　构�?/span>素数p的完全剩余系P={1,2,3,4…�Q�p-1)}�Q�因为（a,p)=1�Q�由引理3可得A={a,2a,3a,4a�Q?#8230;�Q�p-1)a}也是p的一个完全剩余系。��o(h��)W=1*2*3*4…*(p-1�Q�，昄��W≡W(mod p�Q�。��o(h��)Y=a*2a*3a*4a*…�Q�p-1)a�Q�因为{a,2a,3a,4a�Q?#8230;�Q�p-1)a}是p的完全剩余系�Q�由引理2以及(qi��ng)引理4可得a*2a*3a*…�Q�p-1)a≡1*2*3*…�Q�p-1)(mod p�Q�即W*a^(p-1�Q?#8801;W(modp�Q�。易知（W,p)=1�Q�由引理1可知a^(p-1�Q?#8801;1(modp�Q?/span>

luis 2012-04-11 07:16 发表评论

�Ƨ拉定理证明 && �Ƨ拉公式

luis — Tue, 10 Apr 2012 23:11:00 GMT

源地址�Q?/span> http://www.shnenglu.com/zoyi-zhang/articles/43456.html
�Ƨ拉函数 �Q?/span>
�Ƨ拉函数是数��Z��很重要的一个函敎ͼ��Ƨ拉函数是指�Q�对于一个正整数 n �Q�小�?n 且和 n 互质的正整数�Q�包�?1�Q�的个数�Q�记�?φ(n) �?nbsp;

完全余数集合�Q?/span>
定义��于 n 且和 n 互质的数构成的集合�ؓ(f��) Zn �Q�称��D��个集合�ؓ(f��) n 的完全余数集合�?昄�� |Zn| �Q?#966;(n) �?/span>

有关性质�Q?/span>
对于素数 p �Q?#966;(p) = p -1 �?/span>
对于两个不同素数 p�Q?q �Q�它们的乘积 n = p * q 满�� φ(n) = (p -1) * (q -1) �?/span>
�q�是因�ؓ(f��) Zn = {1, 2, 3, ... , n - 1} - {p, 2p, ... , (q - 1) * p} - {q, 2q, ... , (p - 1) * q} �Q?�?φ(n) = (n - 1) - (q - 1) - (p - 1) = (p -1) * (q -1) �Q?/span>φ(p) * φ(q) �?/span>

�Ƨ拉定理 �Q?/span>
对于互质的正整数 a �?n �Q�有 a^φ(n) ≡ 1 mod n �?/span>

证明�Q?br />( 1 ) �?nbsp;Zn = {x₁, x₂, ..., x_φ(n)} �Q?nbsp;S = {a * x₁ mod n, a * x₂ mod n, ... , a * x_φ(n) mod n} �Q?/span>
        �?Zn = S �?/span>
        ① 因�ؓ(f��) a �?n 互质�Q?/span> x_i (1 ≤ i ≤ φ(n)) �?n 互质�Q?所�?a * x_i �?n 互质�Q�所�?a * x_i mod n ∈ Zn �?/span>
        ② �?i ≠ j �Q?那么 x_i ≠ x_j�Q�且�?a, n互质可得 a * x_i mod n ≠ a * x_j mod n �Q�消��d��Q��?/span>

( 2 )     a^φ(n)* x₁* x₂ *... * x_φ(n) mod n
      ≡ (a * x₁) * (a * x₂) * ... * (a * x_φ(n)) mod n
      ≡ (a * x₁ mod n) * (a * x₂mod n) * ... * (a * x_φ(n) mod n) mod n
      ≡  x₁* x₂* ... * x_φ(n)mod n
      �Ҏ(gu��)��{�式的左右两端，因�ؓ(f��) x_i (1 ≤ i ≤ φ(n)) �?n 互质�Q�所�?nbsp;a^φ(n)  ≡ 1 mod n �Q�消��d��Q��?br />注：(x��)
消去律：(x��)如果 gcd(c,p) = 1 �Q�则 ac ≡ bc mod p ⇒ a ≡ b mod p �?br />
贚w��定理 �Q?/span>
若正整数 a 与素�?p 互质�Q�则�?nbsp;a^{p - 1} ≡ 1 mod p �?br />证明�q�个定理非常��单，�׃�� φ(p) = p -1�Q�代入欧拉定理即可证明�?/span>
*****************************************************************************
补充�Q�欧拉函数公�?br />

( 1 ) p^k 的欧拉函�?/h2>

对于�l�定的一个素�?p �Q?φ(p) = p -1。则对于正整�?n = p^k �Q?/p>

 φ(n) = p^k - p^{k -1}
  
证明�Q?/span>
 ���于 p^k 的正整数个数�?p^k - 1个，其中
 �?p^k 不互质的正整数有{p * 1,p * 2,...,p * (p^{k - 1}-1)} ��p�� p^{k - 1} - 1 �?/span>
 所�?φ(n) = p^k - 1 - (p^{k - 1} - 1) = p^k - p^{k - 1} �?/span>

( 2 ) p * q 的欧拉函�?/h2>

假设 p, q是两个互质的正整敎ͼ��?p * q 的欧拉函��Cؓ(f��)

φ(p * q) = φ(p) * φ(q) �Q?gcd(p, q) = 1 �?/p>

证明�Q?/span>
 �?n = p * q �Q?gcd(p,q) = 1
 �Ҏ(gu��)��中国余数定理�Q�有
 Zn �?Zp × Zq 之间存在一一映射
�Q�我的想法是�Q?a ∈ Zp �Q?b ∈ Zq ⇔ b * p + a * q ∈ Zn �?/span>�Q?/span>
 所�?n 的完全余数集合的元素个数�{�于集合 Zp × Zq 的元素个数�?/span>
 而后者的元素个数�?φ(p) * φ(q) �Q�所以有
 φ(p * q) = φ(p) * φ(q) �?/span>

( 3 ) ��L��正整数的�Ƨ拉函数

��L��一个整�?n 都可以表�C�Zؓ(f��)其素因子的乘�U��ؓ(f��)�Q?/p>

      I
 n = ∏  p_i^k_i (I �?n 的素因子的个�?
     i=1

�Ҏ(gu��)��前面两个�l�论�Q�很�Ҏ(gu��)��得出它的�Ƨ拉函数为：(x��)


         I                      I
 Φ(n) = ∏  p_i^k_i-1(p_i-1) = n ∏ (1 - 1 / pi)
        i=1                    i=1

对于��L�� n > 2�Q?span style="line-height: 24px; font-weight: bold; ">2 | Φ(n) ,因�ؓ(f��)必存�?nbsp; p_i-1 是偶数�?/p>

luis 2012-04-11 07:11 发表评论

joj 2296 Boxs �l�合数学

luis — Tue, 21 Jul 2009 04:21:00 GMT

题意是N个不同的盒子�Q�A个完全相同的�U�球�Q�B个完全相同的��球�Q�现在把球往盒里装，球可以不全部装到盒里�Q�盒可以为空�Q�问�Ҏ(gu��)��数�?/p>

其实��是把A个红球，B个篮球分成n+1堆（除n堆外�q�有一堆就是没有放入盒中的�Q��?br>思�\很重要，
我的思�\是由于A球，B球不同色�Q�可以看成两个独立事�Ӟ��则结果�ؓ(f��)count(A)*count(B),
其中count(A)表示A的放法，
则利用隔板法(或者多重集的r�l�合)�Q�count(A)=C(n+1+A-1,n+1-1 ),利用pascal公式c(n,m)=c(n-1,m)+c(n-1,m-1)来求该式;

�q�有一个注意点当且仅当n=20,a= 15 ,b=15时会(x��)��出long long 范围。所以特�D�处�?br>

�?+"展开

代码�?�?x��)自动展开

luis 2009-07-21 12:21 发表评论

joj 1206 Word Index 利用�l�合数学性质推荐

luis — Sat, 18 Jul 2009 02:08:00 GMT

Encoding schemes are often used in situations requiring encryption or information storage/transmission economy. Here, we develop a simple encoding scheme that encodes particular types of words with five or fewer (lower case) letters as integers.

Consider the English alphabet {a,b,c,...,z}. Using this alphabet, a set of valid words are to be formed that are in a strict lexicographic order. In this set of valid words, the successive letters of a word are in a strictly ascending order; that is, later letters in a valid word are always after previous letters with respect to their positions in the alphabet list {a,b,c,...,z}. For example,

abc aep gwz

are all valid three-letter words, whereas

aab are cat

are not.

For each valid word associate an integer which gives the position of the word in the alphabetized list of words. That is:

    a -> 1
b -> 2
.
.
z -> 26
ab -> 27
ac -> 28
.
.
az -> 51
bc -> 52
.
.
vwxyz -> 83681

Your program is to read a series of input lines. Each input line will have a single word on it, that will be from one to five letters long. For each word read, if the word is invalid give the number 0. If the word read is valid, give the word's position index in the above alphabetical list.

Input

The input consists of a series of single words, one per line. The words are at least one letter long and no more that five letters. Only the lower case alphabetic {a,b,...,z} characters will be used as input. The first letter of a word will appear as the first character on an input line.

The input will be terminated by end-of-file.

Output

The output is a single integer, greater than or equal to zero (0) and less than or equal 83681. The first digit of an output value should be the first character on a line. There is one line of output for each input line.

Sample Input

z
a
cat
vwxyz

Sample Output

利用�l�合数学定理
{1,2,3,--,n}的r�l�合a1,a2,--,ar出现在{1�Q?�Q?-�Q�n}的r�l�合的字典序中的位置号如下：(x��)

C(n,r)-C(n-a1,r)-C(n-a2,r-1)-------C(n-ar-1�Q?)-C(n-ar,1):证明略；

�l�合数C(n,r)计算利用pascal公式。C(n+1,k)=C(n,k-1)+C(n,k)

luis 2009-07-18 10:08 发表评论

luis — Sat, 27 Jun 2009 02:44:00 GMT

One curious child has a set of N little bricks. From these bricks he builds different staircases. Staircase consists of steps of different sizes in a strictly descending order. It is not allowed for staircase to have steps equal sizes. Every staircase consists of at least two steps and each step contains at least one brick. Picture gives examples of staircase for N=11 and N=5:

Your task is to write a program that reads from input numbers N and writes to output numbers Q - amount of different staircases that can be built from exactly N bricks.

Input

Numbers N, one on each line. You can assume N is between 3 and 500, both inclusive. A number 0 indicates the end of input.

Output

Numbers Q, one on each line.

Sample Input

3
5
0

Sample Output

1
2

�Ҏ(gu��)��1�Q�动态规�?br>
#include
#include
using namespace std;
int main()
{
freopen("s.txt","r",stdin);
freopen("key.txt","w",stdout);
double f[501][501]={0};
double s;
int i,j,k,n;
for(i=3;i<=500;i++)
for(j=1;j<=(i-1)/2;j++)
    f[i][j]=1;
for(i=3;i<=500;i++)
    for(j=1;j<=(i-1)/2;j++)
    {
for(k=j+1;k<=(i-j-1)/2;k++)
   f[i][j]=f[i-j][k];
}
    while(scanf("%d",&n),n) {
        s=0;
        for(i=1;i<=(n-1)/2;i++) s+=f[n][i]; //f[n]=f[n][1]+f[n][2]+-----+f[n][floor((i-1)/2)]
        printf("%.0f\n",s);
    }
//system("PAUSE");
return   0;
}
更妙的方法：(x��)生成函数�?br>计算(1+x)(1+x^2)(1+x^3)-----,x^n的系数即为所�?br>int i,j;
double ans[510]={1,1};//已经把ans[1]和ans[0]赋�ؓ(f��)1了，其余�?
for(i=2;i<=500;i++) {
        for(j=500;j>=0;j--) {
            if(i+j<=500) ans[i+j]+=ans[j];
        }
    }

先计��?1+x)(1+x^2)
再计��?1+x)(1+x^2)   *(1+x^3)

luis 2009-06-27 10:44 发表评论

<转帖> joj 1978

luis — Tue, 12 May 2009 03:42:00 GMT

#include <cstdio>
#include <string>

int b[51][51][50], N;

void add ( int i, int j )
{
    int k;
    for ( k = 0; k < 50; k ++ )
        b[i][j][k] = b[i - 1][j - 1][k] + b[i - 1][j][k] * j;
    int c = 0, t;
    for ( k = 0; k < 50; k ++ )
    {
        t = b[i][j][k] + c;
        c = t / 10;
        b[i][j][k] = t % 10;
    }
}

void dp ()
{
    memset ( b, 0x00, sizeof ( b ) );
    int i, j;
    for ( i = 1; i <= 50; i ++ )
    {
        b[i][1][0] = b[i][i][0] = 1;
    }
    for ( i = 3; i <= 50; i ++ )
    {
        for ( j = 2; j < i; j ++ )
        {
            add ( i, j );
        }
    }
}

void print ( int i, int j )
{
    int k;
    for ( k = 49; k >= 0; k -- )
        if ( b[i][j][k] )
            break;
    if ( k == -1 )
        printf ( "0" );
    for ( ; k >= 0; k -- )
        printf ( "%d", b[i][j][k] );
    printf ( " " );
}

void print ( int n )
{
    printf ( "%d ", n );
    int i, j, k;
    int ans[50];
    memset ( ans, 0, sizeof ( ans ) );
    for ( i = 1; i <= n; i ++ )
    {
        for ( j = 0; j < 50; j ++ )
        {
            ans[j] += b[n][i][j];
        }
    }
    int t, c = 0;
    for ( k = 0; k < 50; k ++ )
    {
        t = ans[k] + c;
        c = t / 10;
        ans[k] = t % 10;
    }
    for ( k = 49; k >= 0; k -- )
        if ( ans[k] )
            break;
    if ( k == -1 )
        printf ( "0" );
    for ( ; k >= 0; k -- )
        printf ( "%d", ans[k] );
    printf ( " " );
}

int main ()
{
    //freopen ( "in.txt", "r", stdin );
    dp ();
    //print ( 5, 2 );
    while ( scanf ( "%d", &N ) && N )
    {
        print ( N );
    }
    return 0;
}

Rhyme Schemes

Status	In/Out	TIME Limit	MEMORY Limit	Submit Times	Solved Users	JUDGE TYPE
	stdin/stdout	3s	8192K	97	55	Special Test

The rhyme scheme for a poem (or stanza of a longer poem) tells which lines of the poem rhyme with which other lines. For example, a limerick such as


If computers that you build are quantum
Then spies of all factions will want 'em
Our codes will all fail
And they'll read our email
`Til we've crypto that's quantum and daunt 'em
Jennifer and Peter Short(http://www.research.att.com/~shor/notapoet.html)

Has a rhyme scheme of aabba, indicating that the first, second and fifth lines rhyme and the third and fourth lines rhyme.

For a poem or stanza of four lines, there are 15 possible rhyme schemes: aaaa, aaab, aaba, aabb, aabc, abaa, abab, abac, abba, abbb, abbc, abca, abcb, abcc, and abcd.

Write a program to compute the number of rhyme schemes for a poem or stanza of N lines where N is an input value.

Input

Input will consist of a sequence of integers N, one per line, ending with a 0 (zero) to indicate the end of the data. N is the number of lines in a poem.

Output

For each input integer N, your program should output the value of N, followed by a space, followed by the number of rhyme schemes for a poem with N lines as a decimal integer with at least 12 correct significant digits (use double precision floating point for your computations).

Sample Input

Sample Output

1 1
2 2
3 5
4 15
20 51724158235372
30 846749014511809120000000
10 115975

排列�l�合的题目，大致上从��到大顺推即可。设b[i][j]中i表示字符串长度，j表示字符串中用到的字母个敎ͼ�不难推出b[i][j] = b[i-1][j-1] + b[i - 1][j] * j�?br>而我自己的思�\一直是��x��据最后的那一个字母来推，推不出来

luis 2009-05-12 11:42 发表评论

色综合久久天天综线观看,人妻无码久久精品,一级做a爰片久久毛片免费陪

假如p是质敎ͼ��?a,p)=1�Q�那�?a^(p-1) ≡1�Q�mod p�Q? 假如p是质敎ͼ�且a,p互质�Q�那�?a�?p-1)�ơ方除以p的余数恒�{�于1 历史

�~�辑本段证明

�Ƨ拉定理证明 && �Ƨ拉公式

( 3 ) ��L��正整数的�Ƨ拉函数

joj 2296 Boxs �l�合数学

joj 1206 Word Index 利用�l�合数学性质 推荐

Input

Output

Sample Input

Sample Output

Input

Output

Sample Input

Sample Output

<转帖> joj 1978

Input

Output

Sample Input

Sample Output

假如p是质敎ͼ��?a,p)=1�Q�那�?a^(p-1) ≡1�Q�mod p�Q?
假如p是质敎ͼ�且a,p互质�Q�那�?a�?p-1)�ơ方除以p的余数恒�{�于1
历史

joj 1206 Word Index 利用�l�合数学性质推荐