亚洲国产精品无码久久久秋霞2,久久66热人妻偷产精品9,国内精品久久国产大陆

pow函数的性能��试

beyonlin — Sat, 25 Aug 2007 12:16:00 GMT

昨天在PKU上做了一�?a >2187�Q�限�?s�?br>��法主要耗时在多�ơ求不同整数的��^斏V�?br>当用pow函数求时�Q�超�Ӟ��
而直接乘�?32ms�?br>相差也太大了吧�?br>于是��写了一�D�代码来��试pow的性能
首先产生10000个随机整敎ͼ�然后重复1000�ơ求整数的��^�?/p>

#include <iostream>
#include <cmath>
#include <ctime>
using namespace std;
const int MAX = 10000;
int a[MAX];
int main()
{
    int i, j, n = MAX;
    int rep = 1000; //重复�ơ数
    clock_t beg, end;
    for(i = 0; i < n; i++)
        a[i] = rand() % 20000 - 10000; //-10000 <= a[i]< 10000

    cout<<"test a[i]*a[i]"<<endl;
    beg = clock();
    for(j = 0; j < rep; j++)
        for(i = 0; i < n; i++)
            a[i] * a[i];
    end = clock();
    cout<<"time: "<<end - beg<<"ms"<<endl;

    cout<<"test pow(a[i], 2.0)"<<endl;
    beg = clock();
    for(j = 0; j < rep; j++)
        for(i = 0; i < n; i++)
            pow(a[i], 2.0);
    end = clock();
    cout<<"time: "<<end - beg<<"ms"<<endl;

    return 0;
}

下面是测试结果：

test a[i]*a[i]
time: 31ms
test pow(a[i], 2.0)
time: 2828ms

所以下�ơ遇到类似情况不再用pow函数�?#8230;…

beyonlin 2007-08-25 20:16 发表评论

beyonlin — Sun, 12 Aug 2007 16:29:00 GMT

在做PKU2762�Ӟ��需要徏��L��表�?br>于是按部��q��写了下面一个插入边到邻接表中的函数�Q?/p>

const int VMAX = 1010;
typedef struct Graph
{
    int vex;
    Graph* next;
}Graph;
Graph ArcGraph[VMAX];
void insert(int u, int v)
{
    Graph* t = new Graph;
    Graph* p = ArcGraph[u].next;
    t->vex = v;
    t->next = p;
    ArcGraph[u].next = t;
}

完成完整的程序提交上去，得到�l�果
Memory:25796K Time:375MS
Language:C++ Result:Accepted

再对比别人的�E�序
Memory:296K Time:109MS

无论是时间还是空间都相差很大�?br>于是��p��虑怎么优化自己的程序�?br>�W�一个问题：规模只有1000�Q��ؓ什么会用那么多内存呢？
仔细一��x��据是多case的，每次插入新节�Ҏ��都要动态创��Z��个节炏V�?br>一来动态创��时��_��二来每个case�l�束的邻接表中的节点没有释放�Q�故耗费大量内存�?br>然后��想��C��下面的算法，首先初始化一块内存Graph use[100*VMAX];�q�样每次需要新节点�Ӟ��
��׃��use中获取。如果use使用完毕��再动态创建�?br>
依此��法优化后，得到的结果比较满�?br>Memory:1000K Time:218MS
Language:C++ Result:Accepted

const int VMAX = 1010;
typedef struct Graph
{
    int vex;
    Graph* next;
}Graph;
Graph ArcGraph[VMAX];
Graph use[100*VMAX];
int size = 0;
void insert(int u, int v)
{
    Graph* t;
    if(size < 100*VMAX)
    {
        t = &use[size];
        size++;
    }
    else t = new Graph;
    Graph* p = ArcGraph[u].next;
    t->vex = v;
    t->next = p;
    ArcGraph[u].next = t;
}

beyonlin 2007-08-13 00:29 发表评论

beyonlin — Fri, 18 May 2007 08:13:00 GMT

#include<iostream>
using namespace std;
const int MAXV = 10000; //素数表范�?br>bool flag[MAXV+1]; //标志一个数是否为素�?br>int prime[MAXV+1]; //素数�?下标�?开�?br>int size; //素数个数
void genPrime(int max)
{
    memset(flag, true, sizeof(flag));
    for(int i = 2; i <= max / 2; i++)
    {
        if(flag[i])
        {
            for(int j = i << 1 ; j <= max; j += i)
            {
                flag[j] = false;
            }
        }
    }
    for(int i = 2 ; i <= max; i++)
    {
        if(flag[i])
        {
            prime[size++] = i;
        }
    }
}
int main()
{
    genPrime(MAXV);
    return 0;
}

beyonlin 2007-05-18 16:13 发表评论

字符串hash函数

beyonlin — Sat, 07 Apr 2007 08:22:00 GMT

字符串hash函数�Q�解军_��H�用开攑֮�址法，每次对哈希值加1
在下列程序中�Q�不是按常规�Ҏ��用哈希表来记录关键字�Q?br>而是用整型数�l�Htable记录关键字在字符串ch中的位置�?br>在插入时不用把关键字复制到哈希表中，只是记录一个烦引，从而提高了效率�?br>当查询时�Q�只要把Htable的值映��到字符串ch中就可以了�?br>注意ch的下标要�?开始，因�ؓHtable中的零��D��为是�I�，处理��h��比较方便�?

#include<iostream>
#include<string>
using namespace std;
const int MAXN = 9973; //哈希表长�?br>const int len = 30; //字符串的最大长�?br>int Htable[MAX];
char ch[MAX][len]; //存储关键字的字符�?br>unsigned long Hash(char * key)
{
    unsigned long h = 0;
    while(*key)
    {
        h = (h << 4) + *key++;
        unsigned long g = h & 0xf0000000L;
        if(g)
            h ^= g >> 24;
        h &= ~g;
    }
    return h % MAX;
}
int search(char * key)
{
    unsigned long i = Hash(key);
    while(Htable[i])
    {
        if(strcmp(ch[Htable[i]], key) == 0)
            return i;
        i = (i + 1) % MAX;
    }
    return -1;
}
int insert(char * key, int j) //j为关键字在ch中的位置�Q�即索引
{
    unsigned long i = Hash(key);
    while(Htable[i])
        i = (i + 1) % MAX;
    Htable[i] = j;
    return i;
}

beyonlin 2007-04-07 16:22 发表评论

插入排序泛型��法

beyonlin — Tue, 06 Feb 2007 11:13:00 GMT

template 
void insertSort(Iter *begin, Iter *end)
{
	for(Iter *it = begin + 1; it != end; it++)
	{
		Iter tmp = *it;
		Iter *it2 = it - 1;
		while(it2 > begin - 1 && *it2 > tmp)
		{
			*(it2 + 1) = *it2;
			it2 --;
		}
		*(it2 + 1) = tmp;
	}
}

beyonlin 2007-02-06 19:13 发表评论

最大匹配匈牙利��法

beyonlin — Tue, 24 Oct 2006 10:34:00 GMT

#include
#include
using namespace std;
const int MAX = 10001;
bool map[MAX][MAX],visit[MAX];
int match[MAX];
int a, b;
bool DFS(int i)
{
	int j, k = 1;
	for(j = 1 ; j <= b; j++)
	{
		if(map[i][j] && !visit[j])
		{
			visit[j] = true;
			k = match[j];
			match[j] = i;
			if(k == 0 || DFS(k))
				return true;
			match[j] = k;
		}
	}
	return false;
}
int Hungary()
{	
	int	j, ans = 0;
	memset(match, 0, sizeof(match));
	for(j = 1; j <= a; j++)
	{
		memset(visit,false, sizeof(visit));
		if(DFS(j))
			ans++;
	}
	return ans;
}
int main()
{
	int i, j, n,ans;
	memset(map,false,sizeof(map));
	scanf("%d%d",&a,&b);
	ans = Hungary();
	printf("%d\n",ans);	
	return 0;
}

beyonlin 2006-10-24 18:34 发表评论

最��生成树Prim��法

beyonlin — Fri, 13 Oct 2006 15:48:00 GMT

#include
const int MAX = 10000;
const int INF = 1000000;
int clo[MAX];
int low[MAX];
int c[MAX][MAX];
bool flag[MAX];
int beg[MAX],end[MAX];//记录生成树的每条边的两个��点
int Prim(int n)
{
	int i, j, k, ans = 0, pair = 0;
	flag[1] = true;
	for(i = 2; i <= n; i++)
	{
		low[i] = c[1][i];
		clo[i] = 1;
		flag[i] = false;
	}
	for(i = 1; i < n; i++)
	{
		j = 1;
		int min = INF;
		for(k = 2; k <=n; k++)
		{
			if(low[k] < min && !flag[k])
			{
				min = low[k];
				j = k;
			}
		}
		flag[j] = true;
		beg[i] = j;
		end[i] = clo[j];
		ans += c[j][clo[j]];
		for(k = 2; k <= n; k++)
		{
			if(c[j][k] < low[k] && !flag[k])
			{
				low[k] = c[j][k];
				clo[k] = j;
			}
		}
	}
	return ans;
}
int main()
{
	int i, j, n, m;
	scanf("%d%d",&n,&m);
	for(i = 1; i <= n; i++)
	{
		for(j = 1; j <=n; j++)
		{
			c[i][j]=INF;
		}
	}
	return 0;
}

beyonlin 2006-10-13 23:48 发表评论

itoa函数

beyonlin — Wed, 11 Oct 2006 16:59:00 GMT

#include
#include
int main()
{
	int num = 10;
	char str[100];
	itoa(num, str, 2);
	printf("%s\n", str);
	return 0;
}

itoa()函数�?个参敎ͼ��W�一个参数是要�{换的数字�Q�第二个参数是目标字�W�串�Q�第三个参数是�{�U�L��字时所�?的基数。在上例中，转换基数�?0�?0�Q�十�q�制�Q?�Q�二�q�制…�?br />于是惛_��了一个十�q�制转二�q�制的方法：

#include
#include
int main()
{
	int num = 10;
	char str[100];
	int n = atoi(itoa(num, str, 2));
	printf("%d\n",n);
	return 0;
}
先把num转换��Z���q�制的字�W�串�Q�再把该字符串�{换�ؓ整数�?/pre>

beyonlin 2006-10-12 00:59 发表评论

beyonlin — Tue, 10 Oct 2006 16:27:00 GMT

转自�Q?a >http://www.blog.edu.cn/user2/sempr/archives/2006/1142716.shtml
#include

const int MAX = 10000;
int a[MAX],b[MAX];
int change;
void merge(int p, int q, int r)
{
	int i, j = 0;
	int begA = p, endA = q, begB = q+1, endB = r;
    while(begA <= endA && begB <= endB)
	{
		if(a[begA] <= a[begB])
			b[j++] = a[begA++];
		else
		{
			b[j++] = a[begB++];
			change += q - begA + 1;
        }
	}
	while(begA <= endA)
		b[j++] = a[begA++];
    while(begB <= endB)
		b[j++] = a[begB++];
    for(i = 0; i < j; i++)
		a[p+i] = b[i];
}
void mergeSort(int first, int last)
{
	if(first < last)
	{
		int mid = (first + last) / 2;
		mergeSort(first, mid);
        mergeSort(mid+1, last);
        merge(first, mid, last);
    }
}
int main()
{
	return 0;
}

beyonlin 2006-10-11 00:27 发表评论

单源最短�\径Dijkstra��法

beyonlin — Mon, 09 Oct 2006 16:22:00 GMT

d[i]用来保存起始点beg到点i的最短�\径�?br />c[i][j]��的权�?br />如果�?lt;i,j>不存在，则置c[i][j]=INF�?br />path[i]用来保存最短�\径中点i的前一个顶炏V�?br />

#include
const int MAX = 10000;
const int INF = 1000000;
int d[MAX];
int c[MAX][MAX];
bool flag[MAX];
//int path[MAX];
int Dijkstra(int beg, int n)
{
	int i, j, u, tmp;
	for(i = 1; i <= n; i++)
	{
		d[i] = c[beg][i];
		flag[i] = false;
		/*if(d[i] == INF)
			path[i] = 0;
		else
			path[i] = beg;*/
	}
	d[beg] = 0; flag[beg] = true;
	for(i = 1; i <= n; i++)
	{
		tmp = INF; u = beg;
		for(j = 1; j <=n; j++)
		{
			if(!flag[j] && d[j] < tmp)
			{
				u = j; 
				tmp = d[j];
			}
		}
		flag[u] = true;
		for(j  = 1; j <= n; j++)
		{
			if(!flag[j] && c[u][j] < INF)
			{
				if(d[u] + c[u][j] < d[j])
					d[j] = d[u] + c[u][j];
				//path[j] = u;
			}
		}
	}
	return 0;
}
int main()
{	
	return 0;
}

beyonlin 2006-10-10 00:22 发表评论

多源最短�\径floyd_warshall��法

beyonlin — Sun, 08 Oct 2006 16:15:00 GMT

d[i][j]保存�?lt;i,j>的权�?br />如果�?lt;i,j>不存在，则置d[i][j]为INF�?br />

#include
const int MAX=10000;
const int INF=1000000;
int d[MAX][MAX];
int floyd (int n)
{

	for(int k =1 ; k <= n; k++)
	{
		for(int i = 1; i <= n; i++)
		{
			for(int j = 1; j <= n; j++)
			{
				if(d[i][k] + d[k][j] < d[i][j])
					d[i][j] = d[i][k] + d[k][j];
			}
		}
	}
        return 0;
}
int main()
{
	return 0;
}
floyd后，如果d[i][j]>=INF�Q�则点i到点j没有路�?br />else点i到点j的最短�\径长度�ؓd[i][j]�?br />

beyonlin 2006-10-09 00:15 发表评论

数论----判断质数

beyonlin — Sat, 07 Oct 2006 16:40:00 GMT

最�q�发现自己对数论几乎是一�H�不通�?br />是时候开始学了�?br />从零开始…�?br />判断一个数是否��敎ͼ�

bool prime(int a)
{
	for(int i=2;i<=sqrt(a);i++)
	{
		if(a%i==0)
			return false;
	}
	return true;
}

beyonlin 2006-10-08 00:40 发表评论

�q�查集类

beyonlin — Tue, 29 Aug 2006 16:27:00 GMT

参考文章：
http://www.shnenglu.com/qywyh/archive/2006/08/16/11301.html
http://www.programfan.com/blog/article.asp?id=16879

#ifndef UFSET_H
#define UFSET_H
class UFset
{
	public:
		UFset(int);
		void Union(int ,int);
		int Find(int);
		int & operator [] (int i){return parent[i];}
		int size(){return length;}
	private:
		int length;//集合的个�?
		int * parent;
};
UFset::UFset(int len)
{
	length = len;
	parent = new int [length + 1];
	for(int k = 1; k <= length; k++)
		parent[k] = -1;
}
int UFset::Find(int x)
{
	 int i;
	 for(i = x; parent[i] >= 0; i = parent[i]);//搜烦根节�?
	 while(i!=x)//路径压羃
	 {
		  int tmp = parent[x];
		  parent[x] = i;
		  x = tmp;
	 }
	 return i;
}
void UFset::Union(int x,int y)//合�ƈ
{
	int pX = Find(x);
	int pY = Find(y);
	if(pX != pY)
	{
		int tmp = parent[pX] + parent[pY];
		if(parent[pX] > parent[pY])
		{
			parent[pX] = pY;
			parent[pY] = tmp;
		}
		else 
		{
			parent[pY] = pX;
			parent[pX] = tmp;
		}
		length--;
	}
}
#endif

beyonlin 2006-08-30 00:27 发表评论

我的动态规划启蒙题

beyonlin — Mon, 28 Aug 2006 02:31:00 GMT

动态规划即是一个重点，又是一个难炏V�?br />今天�l�于做出了一题像��L��动态规划题�?br />Problem Id:1163  User Id:beyonlin_SCUT
Memory:100K  Time:0MS
Language:C++  Result:Accepted
http://acm.pku.edu.cn/JudgeOnline/problem?id=1163

The Triangle
Time Limit:1000MS Memory Limit:10000K
Total Submit:3415 Accepted:1988

Description

7

3   8

8   1   0

2   7   4   4

4   5   2   6   5


(Figure 1)

Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers passed on a route that starts at the top and ends somewhere on the base. Each step can go either diagonally down to the left or diagonally down to the right.

Input
Your program is to read from standard input. The first line contains one integer N: the number of rows in the triangle. The following N lines describe the data of the triangle. The number of rows in the triangle is > 1 but <= 100. The numbers in the triangle, all integers, are between 0 and 99.

Output
Your program is to write to standard output. The highest sum is written as an integer.

Sample Input

Sample Output
30

分析�Q?br />题意��化�ؓ�Q?br />从第一行开始走到最后一行，每步可以向下走或右下走�?br />所求即��Z��W�一行走到最后一行经�q�的数��d��的最大倹{�?br />令p[][]存储input�?br />5
7
3 8
8 1 0
2 7 4 4
|     \ | \
4 5 2 6 5

如上图，令i��Q�j为列,
d[i][j]��Z��W�一行走到第i行第j列的最大倹{�?br />对于(i,j)�q�个点，它可以从不同方向走来�Q�如�? | '代表从上方走来，' \ '代表从左上方走来�?br />
则动态规则方�E��ؓ�Q?br />                 {     d[i-1][1]+p[i][1]   (j=1)
d[i][j]=Max{     Max( d[i-1][j-1] , d[i-1][j] ) + p[i][j]   (1                  {     d[i-1][i-1]+p[i][i]   (j=i)

�l�果为Max(d[n][j]) , (1<=j<=n)

代码如下�Q?pre>

#include
int p[100][100];
int d[100][100];
int Max(int a,int b)
{return a>b?a:b;}
int main()
{
	int i,n;
	scanf("%d",&n);
	for(i=1;i<=n;i++)
	{
		int j;
		for(j=1;j<=i;j++)
			scanf("%d",p[i]+j);
	}
	d[1][1]=p[1][1];
	for(i=2;i<=n;i++)
	{
		int j;
		d[i][1]=d[i-1][1]+p[i][1];
		for(j=2;j<=i;j++)
			d[i][j]=Max(d[i-1][j-1],d[i-1][j])+p[i][j];
		d[i][i]=d[i-1][i-1]+p[i][i];
	}
	int max=0;
	for(i=1;i<=n;i++)
	{
		if(d[n][i]>max)
			max=d[n][i];
	}
	printf("%d\n",max);
	return 0;
}

beyonlin 2006-08-28 10:31 发表评论

深度优先搜烦

beyonlin — Sun, 27 Aug 2006 17:23:00 GMT

今天做出了第一题深度优先搜索题�?br />��x��对广度和深度有了一个基本的了解�?br />学ACM�ȝ��学到了一炚w��暴力解决问题的方法�?br />Problem Id:1154  User Id:beyonlin_SCUT
Memory:32K  Time:155MS
Language:C++  Result:Accepted
http://acm.pku.edu.cn/JudgeOnline/problem?id=1154

LETTERS
Time Limit:1000MS Memory Limit:10000K
Total Submit:694 Accepted:334

Description
A single-player game is played on a rectangular board divided in R rows and C columns. There is a single uppercase
letter (A-Z) written in every position in the board.
Before the begging of the game there is a figure in the upper-left corner of the board (first row, first column). In every move, a player can move the figure to the one of the adjacent positions (up, down,left or right). Only constraint is that
a figure cannot visit a position marked with the same letter twice.
The goal of the game is to play as many moves as possible.
Write a program that will calculate the maximal number of positions in the board the figure can visit in a single game.

Input
The first line of the input contains two integers R and C, separated by a single blank character, 1 <= R, S <= 20.
The following R lines contain S characters each. Each line represents one row in the board.

Output
The first and only line of the output should contain the maximal number of position in the board the figure can visit.

Sample Input

3 6
HFDFFB
AJHGDH
DGAGEH

Sample Output

我的�E�序�Q?br />#include
#include
using namespace std;
struct node
{
	int row;
	int col;
	int dire;
};
char p[30][30];
char flag[30];
int incr[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
int main()
{
	int i,row,col;
	scanf("%d%d",&row,&col);
	getchar();
	char ch[30];
	for(i=1;i<=row;i++)
	{
		gets(ch);
		int j;
		for(j=1;j<=col;j++)
			p[i][j]=ch[j-1];
	}
	//初始化，外加一�?
	for(i=0;i<=col+1;i++)
	{
		p[0][i]='0';
		p[row+1][i]='0';
	}
	for(i=0;i<=row+1;i++)
	{
		p[i][0]='0';
		p[i][col+1]='0';
	}
	int Maxmove=0;//最大步�?
	stackpath;
        //栈初始化
	int r=1,c=1,dire=0,f=0,move=1;
	node in;
	in.row=r;
	in.col=c;
	in.dire=dire;
	path.push(in);
	flag[f++]=p[r][c];
	while(!path.empty())
	{
		if(dire<4)
		{
			int r2=r+incr[dire][0];
			int c2=c+incr[dire][1];
			bool b=true;
			for(int k=0;kMaxmove)
				Maxmove=move;
			move--;
			dire=path.top().dire+1;	//回溯�Q�去除访问标�?
			path.pop();
			flag[--f]='\0';
			if(!path.empty())
			{
				r=path.top().row;
				c=path.top().col;
			}
		}
	}
	printf("%d\n",Maxmove);
	return 0;
}

beyonlin 2006-08-28 01:23 发表评论

�q�度优先搜烦

beyonlin — Thu, 24 Aug 2006 02:25:00 GMT

今天在PKU上做了我�W�一题广度优先搜索题�Q?br />Problem Id:2627  User Id:beyonlin_SCUT
Memory:64K  Time:575MS
Language:C++  Result:Accepted
个�h认�ؓ��法复杂度应该�ؓO(n^2)或更��。不知是不是�q�样�?br />http://acm.pku.edu.cn/JudgeOnline/problem?id=2627

Gopher and hawks
Time Limit:1000MS Memory Limit:65536K
Total Submit:900 Accepted:328

Description
A gopher sits in a hole located at (xs, ys) and wants to get to a hole located at (xt, yt). The gopher can run at a constant speed of v m/sec. However, if the gopher is outside of a hole for more than a m minutes he will become a supper to hawks flying over the holes. Can the gopher make it?

Input
The first line of input contains two positive integer numbers: v -- gopher's speed in meters per second and m -- the time after which the gopher becomes prey to hawks if he stays outside a hole. The second line of input contains two floating point numbers: the (xs,ys) coordinates of the gopher starting hole. The third line contains the (xt, yt) coordinates of the target hole. Each Subsequent line of input contains two floating point numbers: the (x,y) coordinates of a gopher hole. All distances are in metres, to the nearest mm.

Output
If the gopher can make it to the target hole, the output line should read "Yes, visiting n other holes.", where n is the minimal number of intermediate holes the gopher has to visit. If the gopher cannot make it the output line should read "No." There are not more than 1000 gopher holes and all coordinates are between -10000 and +10000.

Sample Input

3 1
0.000 0.000
500.000 0.000
179.000 0.000
358.000 0.000

Sample Output

Yes, visiting 2 other holes.

Hint
Sample input 2
5 1
0.000 0.000
0.000 550.000
179.000 0.000
0.000 301.000

Output for sample input 2
No.

我的�E�序�Q?br />
#include
#include
#include
using namespace std;
struct node
{
	int point;
	int step;
};
double x[1100],y[1100];
bool flag[1100]={false};
int main()
{
	int i,v,t;
	scanf("%d%d",&v,&t);
	t*=60;
	double beginX,beginY,endX,endY;
	scanf("%lf%lf%lf%lf",&beginX,&beginY,&endX,&endY);
	int n=1;
	while(scanf("%lf%lf",x+n,y+n)!=EOF)
		n++;
	x[0]=beginX;
	y[0]=beginY;
	x[n]=endX;
	y[n]=endY;

	node n1;//队列初始�?
	n1.point=0;
	n1.step=0;
	queue que;
	que.push(n1);

	int steps=0;
	while(true)
	{
		if(que.empty())
			break;
		node tmp=que.front();
		que.pop();//出队�?
		for(i=1;i<=n;i++)
		{
			if(!flag[i])//标志是否�q�过队列
			{				
				double time=sqrt(pow(x[i]-x[tmp.point],2.0)+pow(y[i]-y[tmp.point],2.0))/v;
				if(time
				
		


beyonlin 2006-08-24 10:25 发表评论

考试了，ACM暂告一�D�落

beyonlin — Thu, 25 May 2006 12:35:00 GMT

马上��p��期末考了�Q?br />
ACM暂告一�D�落�Q?br />
考完试后再�ȝ��一下这个学期的ACM�?/font>

beyonlin 2006-05-25 20:35 发表评论

My Status

beyonlin — Mon, 17 Apr 2006 04:59:00 GMT

My Status

Problem	status	time
A	Yes	39
B	Yes	77
G	No-Time-limit Exceeded	96
G	Yes	113
C	Yes	211
E	No-Time-limit Exceeded	265

Solved 4 Time 460

beyonlin 2006-04-17 12:59 发表评论

Rank Name Solved Time

beyonlin — Sun, 16 Apr 2006 17:25:00 GMT

Rank Name Solved Time

1 16 - team16 8 984
2 13 - team13 8 1141
3 44 - team44 7 718
4 64 - team64 7 1247
5 65 - team65 6 564
6 10 - team10 6 642
7 21 - team21 6 684
8 7 - team7 6 830
9 89 - team89 5 414
10 73 - team73 5 448
11 83 - team83 5 559
12 18 - team18 5 561
13 24 - team24 5 716
14 51 - team51 5 753
15 81 - team81 5 846
16 56 - team56 5 898
17 15 - team15 4 238
18 72 - team72 4 284
19 57 - team57 4 295
20 49 - team49 4 306
21 43 - team43 4 311
22 62 - team62 4 315
23 4 - team4 4 318
24 8 - team8 4 321
25 26 - team26 4 375
26 3 - team3 4 412
27 45 - team45 4 432
28 69 - team69 4 448
29 27 - team27 4 460
30 88 - team88 4 478
31 50 - team50 4 482
32 66 - team66 4 490
33 48 - team48 4 496
34 11 - team11 4 498
35 79 - team79 4 511
36 71 - team71 4 518
37 47 - team47 4 575
38 77 - team77 4 580
39 19 - team19 4 597
40 9 - team9 4 600
41 68 - team68 4 630
42 90 - team90 4 642
43 20 - team20 4 721
44 37 - team37 4 727
45 29 - team29 4 770
46 12 - team12 4 772
47 61 - team61 4 783
48 22 - team22 4 792
49 41 - team41 4 858
50 63 - team63 4 893
51 34 - team34 4 943
52 76 - team76 3 257
53 54 - team54 3 262
54 39 - team39 3 277
55 60 - team60 3 316
56 82 - team82 3 325
57 1 - team1 3 331
58 25 - team25 3 333
59 36 - team36 3 341
60 86 - team86 3 342
61 70 - team70 3 360
62 52 - team52 3 369
63 78 - team78 3 402
64 14 - team14 3 490
65 6 - team6 3 547
66 80 - team80 3 618
67 5 - team5 2 133
68 38 - team38 2 243
69 2 - team2 2 250
70 74 - team74 2 253
71 55 - team55 2 258
72 58 - team58 2 262
73 84 - team84 2 296
74 23 - team23 2 319
75 40 - team40 2 328
76 91 - team91 2 343
77 42 - team42 2 356
78 100 - other 2 437
79 87 - team87 2 459
80 67 - team67 1 88
81 33 - team33 1 118
82 28 - team28 1 193
83 99 - team99 1 313
84 17 - team17 0 0
84 30 - team30 0 0
84 31 - team31 0 0
84 32 - team32 0 0
84 35 - team35 0 0
84 46 - team46 0 0
84 53 - team53 0 0
84 59 - team59 0 0
84 75 - team75 0 0
84 85 - team85 0 0

我的队号�Q?7-team27

beyonlin 2006-04-17 01:25 发表评论

ACM竞赛落下帷幕

beyonlin — Sun, 16 Apr 2006 17:24:00 GMT

我校�W�二届ACM/ICPC竞赛�l�历5个小时后�l�于落下帷幕�?br />
从早�?0点到下午3点，我队�l�过共同努力�l�于解决�?题中�?题�?br />
�?0多队中排�?9�?br />
因�ؓ是大一�Q�也没有�l�验�Q�又没有看算法书�Q�所以只做出了几题简单的�?br />
也有些大一的牛人搞�?�?题的�Q�真是惭�?

以后�l�箋努力�?br />
先写下我的队名吧�Q�SCTCS

队号�Q?7-team27

有对�U�感吧�?br />
先把各队的成�l�记下来吧�?br />
因队较多�Q�请点击�q�里查看�?/font>

我的比赛状�?/a>�?/font>

beyonlin 2006-04-17 01:24 发表评论

beyonlin — Mon, 27 Mar 2006 17:11:00 GMT

今天做ACM��是�׃��没有注意�q�个问题而Wrong Aswer了几�ơ�?/font>

题目说明要输入的字符串长度最多�ؓ8�?/font>

因此我就只申请了8个字�W��?/font>

�l�果当然��是错了�?/font>

其实字符数组需要一个存储单元来保存�l�束�W�“\0”，

用来说明一个字�W�串�l�束了�?/font>

因此要申��h��字符串多一个存储单元的长度�?/font>

��C��了！�Q?/font>

从此不能犯同样错误！

beyonlin 2006-03-28 01:11 发表评论

beyonlin — Sun, 26 Mar 2006 08:26:00 GMT

做ACM题目已经有两个月了�?br />
在北大的�|�站上也做了几十道题了�?br />
今天在北大网站上又迎来ACM月赛�?br />
旉��是今天中�?2�Q?0�?7:00�?br />
可是做了我三个半��时也没做出来一题来�?br />
不爽�Q�！

beyonlin 2006-03-26 16:26 发表评论

亚洲国产精品无码久久久秋霞2,久久66热人妻偷产精品9,国内精品久久国产大陆

pow函数的性能���试

字符串hash函数

插入排序泛型���法

最大匹配匈牙利���法

最���生成树Prim���法

itoa函数

单源最短�\径Dijkstra���法

多源最短�\径floyd_warshall���法

数论----判断质数

�q�查集类

我的动态规划启蒙题

深度优先搜烦

�q�度优先搜烦

考试了，ACM暂告一�D�落

My Status

Rank Name Solved Time

ACM竞赛落下帷幕

pow函数的性能��试

插入排序泛型��法

最大匹配匈牙利��法

最��生成树Prim��法

单源最短�\径Dijkstra��法

多源最短�\径floyd_warshall��法