组合数学

卑微的网课党来写博客了。。。

今天也是在家宅着的一天。

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鸽巢原理（抽屉原理）

定理: 若有n个鸽巢，n+1只鸽子，则至少有一个鸽巢里至少有两只鸽子。

举个栗子：一年有365天，则在366个人中，一定有至少2人生日在同一天。

一个小应用：在边长为1的正方形内任取5点,则其中至少有2点的距离不超过二分之根号二。

小例题  吃糖果

http://acm.hdu.edu.cn/showproblem.php?pid=1205

是一道典型的鸽巢原理题。可用隔板法进行分析。把拥有最多糖果数量n的一种糖果看成n个挡板，其余所有糖果数为s。

如果s<n-1，则将s个糖果放入挡板时，必定有两个挡板间没有糖果，也就是不能吃完糖，无解。

反之则有解。

#include<bits/stdc++.h>
using namespace std;
int a[1000002];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,max=0,sum=0;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            if(max<a[i]) max=a[i];
        }
        int i=1;
        for(i=1;i<=n;i++)
        {
            if(a[i]!=max) sum+=a[i];
            if(sum>=max-1) break;
        }
        if(i==n+1) printf("No
");
        else printf("Yes
");
    }
    return 0;
}

鸽巢定理加强形式

条件：鸽巢n个，鸽子m₁+m₂+...m_n-n+1只，其中m₁，m₂，...，m_n及n都是正整数。

结论：鸽巢1鸽子数>=m₁, 或鸽巢2鸽子数>=m₂,...或鸽巢n鸽子数>=m_n,至少有一个成立。

Erdös-Szekeres定理

定理: 在由n2+1个实数构成的序列中, 必然含有长为n+1的单调(增或减)子序列。

Ramsey定理

Ramsey定理：对于一个给定的两个整m,n>=2,则一定存在一个最小整数r,使得用两种颜色（例如红蓝）无论给Kr的每条边如何染色，总能找到一个红色的Km或者蓝色的Kn。

K表示完全图，即所有顶点间两两相连构成的图。

r可以看做一个有关m,n的二元函数，即r(m,n)。r(3,3)=6。

基本性质：

①等价性 r(m,n)=r(n,m)

②r(2,n)=n k2较特殊 只有一条边 最小的kr为Kn

③r(m,2)=m

部分Ramsey数表：

 一些题目

Halloween treats

http://acm.hdu.edu.cn/showproblem.php?pid=1808

Description

Every year there is the same problem at Halloween: Each neighbour is only willing to give a certain total number of sweets on that day, no matter how many children call on him, so it may happen that a child will get nothing if it is too late. To avoid conflicts, the children have decided they will put all sweets together and then divide them evenly among themselves. From last year's experience of Halloween they know how many sweets they get from each neighbour. Since they care more about justice than about the number of sweets they get, they want to select a subset of the neighbours to visit, so that in sharing every child receives the same number of sweets. They will not be satisfied if they have any sweets left which cannot be divided.

Your job is to help the children and present a solution.

Input

The input contains several test cases.
The first line of each test case contains two integers c and n (1 ≤ c ≤ n ≤ 100000), the number of children and the number of neighbours, respectively. The next line contains n space separated integers a1 , ... , an (1 ≤ ai ≤ 100000 ), where ai represents the number of sweets the children get if they visit neighbour i.

The last test case is followed by two zeros.

Output

For each test case output one line with the indices of the neighbours the children should select (here, index i corresponds to neighbour i who gives a total number of ai sweets). If there is no solution where each child gets at least one sweet print "no sweets" instead. Note that if there are several solutions where each child gets at least one sweet, you may print any of them.

Sample Input

4 5
1 2 3 7 5
3 6
7 11 2 5 13 17
0 0
Sample Output

3 5
2 3 4

鸽巢原理的应用，大意是在n个数任取c个且是c的倍数的一种取法，输出取的元素的下标。

#include<bits/stdc++.h>
long long c[100005],vis[100005],x;
int main()
{
    int m,n,i,j,flag,t;
    while(scanf("%d%d",&m,&n)&&n+m)
    {
        memset(vis,0,sizeof(vis));
        for(c[0]=flag=0,i=1;i<=n;i++)
        {
            scanf("%I64d",&x);
            if(flag) continue;
            c[i]=c[i-1]+x;
            t=c[i]%m;
            if(t==0)
            {
                for(j=1;j<i;j++) printf("%d ",j);
                printf("%d
",j);
                flag=1;
            }
            else if(vis[t])
            {
                for(j=vis[t]+1;j<i;j++) printf("%d ",j);
                printf("%d
",j);
                flag=1;
            }
            vis[t]=i;
        }
    }
    return 0;
}

Friend-Graph

http://acm.hdu.edu.cn/showproblem.php?pid=6152

Problem Description

It is well known that small groups are not conducive of the development of a team. Therefore, there shouldn’t be any small groups in a good team.
In a team with n members,if there are three or more members are not friends with each other or there are three or more members who are friends with each other. The team meeting the above conditions can be called a bad team.Otherwise,the team is a good team.
A company is going to make an assessment of each team in this company. We have known the team with n members and all the friend relationship among these n individuals. Please judge whether it is a good team.
Input

The first line of the input gives the number of test cases T; T test cases follow.（T<=15）
The first line od each case should contain one integers n, representing the number of people of the team.(n<=3000)

Then there are n-1 rows. The th row should contain n-i numbers, in which number  represents the relationship between member i and member j+i. 0 means these two individuals are not friends. 1 means these two individuals are friends.
Output

Please output ”Great Team!” if this team is a good team, otherwise please output “Bad Team!”.
Sample Input

1 4 1 1 0 0 0 1
Sample Output

Great Team!

用Ramsey定理可以做。

#include<bits/stdc++.h>
using namespace std;
int a[10][10]; 
int main() 
{
    int T;
    scanf("%d",&T);
    while(T--) 
    {
        int n;
        scanf("%d",&n);
        memset(a,0,sizeof(a));
        for(int i=1;i<n;i++)
        {
            for(int j=i+1;j<=n;j++)
            {
                int t;
                scanf("%d",&t);
                if(t&&n<6) a[i][j]=a[j][i]=1;
            }
        } 
        if(n>=6)
        {
            puts("Bad Team!");
            continue;
        }
        int f=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=i+1;j<=n;j++)
            {
                for(int k=j+1;k<=n;++k)
                {
                    if(a[i][j]&&a[i][k]&&a[j][k])
                    {
                        f=1;
                        break;
                    }
                }  
            }
        }
        if(f) puts("Bad Team!");
        else puts("Great Team!");
    }
    return 0;
}

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排列与组合

排列

定义：从m个不同的元素中取出n(0 <= n <= m)个元素排成一列的方法数。用符号A表示。

A(m,n)=m!/(m-n)!

组合

定义：从m个不同的元素中取出n (0 <= n <= m)个元素的方法数。用符号C表示。

A(m,n)=C(m,n)*A(n,n)

C(m,n)=m!/((m-n)!*n!)

杨辉三角和二项式系数

杨辉三角大家应该都知道的，规律也很显然，每一个数由上一行对应的两个数之和。

 

观察(1+x)n的展开形式:

(1+x)⁰ = 1

(1+x)¹ = 1+x

(1+x)² = 1+2x+x²

(1+x)³ = 1+3x+3x²+x³

每一行展开的系数刚好对应杨辉三角每一行的数字。所以，杨辉三角可以用(1+x)ⁿ来定义和计算。

例题

杨辉三角

http://acm.hdu.edu.cn/showproblem.php?pid=2032

#include<bits/stdc++.h>
using namespace std;
int a[31][31];
int main() 
{    
    int i,j;    
    for(i=1;i<=30;i++)    
    {        
        a[i][1]=1;        
        for(j=2;j<=i;j++)        
        {            
            a[i][j]=a[i-1][j-1]+a[i-1][j];        
        }
    }    
    int n;    
    while(~scanf("%d",&n))    
    {      
        for(i=1;i<=n;i++)      
        {
              printf("%d",a[i][1]);
              for(j=2;j<=i;j++)    
              printf(" %d",a[i][j]);          
            printf("
");      
        }      
        printf("
");    
    }    
    return 0;
}

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我讨厌死机的电脑。

为什么每次写博客我的电脑都会重新来过呢。[手动再见]