Daily Archives: 8 3 月, 2013

POJ 1091(跳蚤-扩展欧几里德有解推论+容斥原理)

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Language:
跳蚤
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 6911   Accepted: 1951

Description

Z城市居住着很多只跳蚤。在Z城市周六生活频道有一个娱乐节目。一只跳蚤将被请上一个高空钢丝的正中央。钢丝很长，可以看作是无限长。节目主持人会给该跳蚤发一张卡片。卡片上写有N+1个自然数。其中最后一个是M，而前N个数都不超过M，卡片上允许有相同的数字。跳蚤每次可以从卡片上任意选择一个自然数S，然后向左，或向右跳S个单位长度。而他最终的任务是跳到距离他左边一个单位长度的地方，并捡起位于那里的礼物。 
比如当N=2，M=18时，持有卡片(10, 15, 18)的跳蚤，就可以完成任务：他可以先向左跳10个单位长度，然后再连向左跳3次，每次15个单位长度，最后再向右连跳3次，每次18个单位长度。而持有卡片(12, 15, 18)的跳蚤，则怎么也不可能跳到距他左边一个单位长度的地方。 
当确定N和M后，显然一共有M^N张不同的卡片。现在的问题是，在这所有的卡片中，有多少张可以完成任务。 

Input

两个整数N和M(N <= 15 , M <= 100000000)。

Output

可以完成任务的卡片数。

Sample Input

2 4

Sample Output

12

Hint

这12张卡片分别是： 
(1, 1, 4), (1, 2, 4), (1, 3, 4), (1, 4, 4), (2, 1, 4), (2, 3, 4), 
(3, 1, 4), (3, 2, 4), (3, 3, 4), (3, 4, 4), (4, 1, 4), (4, 3, 4) 

Source

HNOI 2001

假设

a1x1+a2x2+--anxn+a(n+1)M=1

则(a1,a2,..an,M)=1

所以用容斥原理计算出gcd≠1的情况的总数

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<functional>
using namespace std;
#define MAXN (15)
#define MAXM (100000000)
int n,m;
long long ans=0,temp=0;
int a[MAXM],tot=0;
long long pow2(long long a,int b)
{
	if (b==1) return a;
	else if (b%2)
	{
		long long p=pow2(a,b/2);
		return p*p*a;
	}
	else
	{
		long long p=pow2(a,b/2);
		return p*p;
	}
}
void dfs(int j,int c,int cost,int l)
{
	if (c==l)
	{
		int p=m/cost;
		if (l&1) ans+=pow2(p,n);
		else ans-=pow2(p,n);
		return;
	}
	for (int i=j+1;i<=tot+1-l+c;i++)
		dfs(i,c+1,cost*a[i],l);



}
int main()
{
	scanf("%d%d",&n,&m);
	int m2=m;
	for (int i=2;i<=(int)sqrt((double)m2);i++)
		if (0==m2%i)
		{
			a[++tot]=i;
			while (0==m2%i) m2/=i;
		}
	if (m2^1) a[++tot]=m2;
//	for (int i=1;i<=tot;i++) cout<<a[i]<<' ';
	for (int i=1;i<=tot;i++) dfs(0,0,1,i);
//	ans=pow(m,n)-ans;
	printf("%lldn",(long long)pow2(m,n)-ans);


	return 0;
}

POJ 1330(Nearest Common Ancestors-Lca模板题)

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Nearest Common Ancestors
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 13559   Accepted: 7236

Description

A rooted tree is a well-known data structure in computer science and engineering. An example is shown below: 

 
In the figure, each node is labeled with an integer from {1, 2,...,16}. Node 8 is the root of the tree. Node x is an ancestor of node y if node x is in the path between the root and node y. For example, node 4 is an ancestor of node 16. Node 10 is also an ancestor
of node 16. As a matter of fact, nodes 8, 4, 10, and 16 are the ancestors of node 16. Remember that a node is an ancestor of itself. Nodes 8, 4, 6, and 7 are the ancestors of node 7. A node x is called a common ancestor of two different nodes y and z if node
x is an ancestor of node y and an ancestor of node z. Thus, nodes 8 and 4 are the common ancestors of nodes 16 and 7. A node x is called the nearest common ancestor of nodes y and z if x is a common ancestor of y and z and nearest to y and z among their common
ancestors. Hence, the nearest common ancestor of nodes 16 and 7 is node 4. Node 4 is nearer to nodes 16 and 7 than node 8 is. 

For other examples, the nearest common ancestor of nodes 2 and 3 is node 10, the nearest common ancestor of nodes 6 and 13 is node 8, and the nearest common ancestor of nodes 4 and 12 is node 4. In the last example, if y is an ancestor of z, then the nearest
common ancestor of y and z is y. 

Write a program that finds the nearest common ancestor of two distinct nodes in a tree. 

Input

The input consists of T test cases. The number of test cases (T) is given in the first line of the input file. Each test case starts with a line containing an integer N , the number of nodes in a tree, 2<=N<=10,000. The nodes are labeled with integers 1, 2,...,
N. Each of the next N -1 lines contains a pair of integers that represent an edge --the first integer is the parent node of the second integer. Note that a tree with N nodes has exactly N - 1 edges. The last line of each test case contains two distinct integers
whose nearest common ancestor is to be computed.

Output

Print exactly one line for each test case. The line should contain the integer that is the nearest common ancestor.

Sample Input

2
16
1 14
8 5
10 16
5 9
4 6
8 4
4 10
1 13
6 15
10 11
6 7
10 2
16 3
8 1
16 12
16 7
5
2 3
3 4
3 1
1 5
3 5

Sample Output

4
3

Source

Taejon 2002


Rt,纯粹无聊写的……(其实至今为止就写过2个………("▔□▔)/)

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define MAXN (10000+10)
#define Li (17)
int t,n,father[MAXN][Li];
int edge[MAXN],pre[MAXN],next[MAXN],size=0;
void addedge(int u,int v)
{
	edge[++size]=v;
	next[size]=pre[u];
	pre[u]=size;
}

int queue[MAXN],deep[MAXN],bin[Li];
void bfs()
{
	int head=1,tail=1;
	for (int i=1;i<=n;i++)
		if (!father[i][0]) {queue[1]=i;deep[i]=1;break;}
	while (head<=tail)
	{
		int &u=queue[head];
		for (int i=1;i<Li;i++)
		{
			father[u][i]=father[father[u][i-1]][i-1];
			if (father[u][i]==0) break;
		}
		for (int p=pre[u];p;p=next[p])
		{
			int &v=edge[p];
			deep[v]=deep[u]+1;
			queue[++tail]=v;
		}
		head++;
	}
}
int lca(int x,int y)
{
	if (x==y) return x;
	if (deep[x]<deep[y]) swap(x,y);
	int t=deep[x]-deep[y];
	for (int i=0;i<Li;i++)
		if (bin[i]&t)
		{
			x=father[x][i];
		}
	int i=Li-1;
	while (x^y)
	{
		while (father[x][i]==father[y][i]&&i) i--;
		x=father[x][i];y=father[y][i];
	}
	return x;
}
int main()
{
//	freopen("poj1330.in","r",stdin);
	scanf("%d",&t);
	for (int i=0;i<Li;i++) bin[i]=1<<i;

	while (scanf("%d",&n)!=EOF)
	{
		memset(pre,0,sizeof(pre));size=0;
		memset(father,0,sizeof(father));
		for (int i=1;i<n;i++)
		{
			int u,v;
			scanf("%d%d",&u,&v);
			father[v][0]=u;
			addedge(u,v);
		}
		bfs();
		int x,y;
		scanf("%d%d",&x,&y);
		printf("%dn",lca(x,y));

	}
	return 0;
}