其他

The Power of XiangqiTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 13    Accepted Submission(s): 6Problem DescriptionXiangqi is one of the most popular two-player board games in China. The

C++的文檔中說，STL中的unique是類似於這樣實現的：template <class ForwardIterator> ForwardIterator unique ( ForwardIterator first, ForwardIterator last ){ ForwardIterator result=first; while (++first != last) { if (!(*result == *first)) // or: if

MysteryTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 141    Accepted Submission(s): 75 InputThe first line of input contains a single integer P, (1 <= P <= 1000), which is the number

The New VillaMr. Black recently bought a villa in the countryside. Only one thing bothers him: although there are light switches in most rooms, the lights they control are often in other rooms than the switches themselves. While his estate agent saw

View Code #include<stdio.h>#include<string.h>#define min(a,b)(a<b?a:b)#define INF 9999999#define MAX 1005int S,N,end,M;int cap[MAX][MAX],flow[MAX],p[MAX];int room[MAX],belong[MAX];int q[10000];int bfs(){int u,i;for(i=0;i<MAX;i++)

每次放考慮有空盤的情況和沒空盤的情況，由於較小的m、n已經打好表，所以可直接調用。自底向上dp。#include <iostream>using namespace std; long a[20][20]; int main(){         long t,n,m,i,j;         for(i=0;i<=10;i++)          {a[0][i]=1; a[i][0]=0;}         for(i=1;i<=10;i++)         {  

A Magic LampTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 745    Accepted Submission(s): 271Problem DescriptionKiki likes traveling. One day she finds a magic lamp, unfortunately the genie in

#include<stdio.h>#include<math.h>#include<algorithm>#include<iostream>using namespace std;const int MAXN=1000;struct point{ int x,y;};point list[MAXN];int stack[MAXN],top;int cross(point p0,point p1,point p2) //計算叉積 p0p1

Super MarioTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 120    Accepted Submission(s): 73Problem DescriptionMario is world-famous plumber. His “burly” figure and amazing jumping ability

題目來源：POJ 1426 Find The Multiplehttp://acm.pku.edu.cn/JudgeOnline/problem?id=1426 解法類型：深度優先搜尋 作者：劉亞寧 題目大意：找出任意一個n(n<=100)的倍數中只由01構成的十進位數。 解題思路：以一個1為根節點，每個節點接有0和1構建一棵樹，應用深度優先搜尋，得到符合題意的解。 提交情況：1、Compile Error一次：編譯器選擇問題，選擇G++編譯器後此問題解決。                

題意：有一個售貨員，n個顧客，賣給 i 這個顧客的價格是pi，i 顧客來買東西的時刻為 ti如果有w的時間段沒人來買東西，售貨員就會睡著，下一個顧客來時會叫醒她，但是不買東西售貨員會賣給第i個來（相對順序）的顧客 1 + ((i - 1) mod 3)的數量的麵包 即 1 2 3 中的一個問使得平均的銷售額最大的最小w以及此時的平均銷售額是多少（

ZOJ Problem Set - 3203Light BulbTime Limit: 1 Second      Memory Limit: 32768 KBCompared to wildleopard's wealthiness, his brother mildleopard is rather poor. His house is narrow and he has only one light bulb in his house. Every night, he is

以下內容轉自 http://www.cnblogs.com/goodness/admin/www.sjchaxun.cn/http/ ，因為網頁上不去，所以將Google cache中的東西拷貝了下來。    當使用者點擊或搜尋引擎向網站伺服器發出瀏覽請求時，伺服器將返回Http Header Http頭資訊狀態代碼,常見幾種如下:1、Http/1.1 200 OK 訪問正常  表示成功訪問,為網站可正常訪問時的狀態。2、Http/1.1 301 Moved Permanently

最近才開始寫圖形化的程式。不禁感慨道雖然也有很多不足，Windows作為一個作業系統是多麼的強大！當然，這也導致寫程式很麻煩，尤其是關係到unicode編碼的時候。今天遇到的問題是關於GDI對象的，代碼如下： 1 void Cls_OnPaint(HWND hDlg) 2 { 3 PAINTSTRUCT ps; 4 HDC hDc, memDc; 5 HBITMAP hOldBitmap; 6 7 hDc = BeginPaint(hDlg, &ps);

80ers' MemoryTime Limit: 1 Second      Memory Limit: 32768 KBI guess most of us are so called 80ers, which means that we were born in the 1980's. This group of people shared a lot of common memories. For example, the Saint Seiya, the YoYo ball, the

有一個連通器，由兩個漏鬥組成，倒vol提及的水，問你連通器中水的高度是多少解法：先求出水面可以到達的最低和最高的高度，然後二分答案判斷View Code #include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;const double eps = 1e-8;const int maxn = 1010;const double

http://codeforces.com/problemset/problem/12/D線段樹View Code #include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int inf = ~0u>>2;const int N = 500010;#define lson l,m,rt<<1#define rson

用位元運算來實現廣搜，每一條邊就是一個方向View Code #include<stdio.h>#include<queue>#include<string.h>using namespace std;int steps;queue<int> Q;int vis[1<<17];int dir[32][2]={0,1,0,2,0,8,0,4, 1,3,1,9,1,5, 3,7,3,2,3,11, 11,9,11,15,11,10, 10

一覺醒來，不覺已經是4點了，於是果斷挑了一道水題。View Code #include<stdio.h>#include<string.h>#include <cctype>char str[5000];int main(){int i,j,k;while(gets(str)) {char s[]="0000";int len=strlen(str);int row=len/16;int left=len%16;int

題目資料較小，直接枚舉自己和敵人的所有全排列，暴力水之（討論出做法後，隊友寫的）View Code #include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int n;int id1[10],id2[10];struct node{char s[25];}p[10];int map[10][10];int cmp(node a,node b){return