其他

#include <iostream>#include <cstring>#include <cstdio>#include <cstdlib>#include <queue>using namespace std;/** Reference:http://www.cnblogs.com/procedure2012/archive/2012/01/29/2331460.html Thanks to:

A. Flipping Gametime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputIahub got bored, so he invented a game to be played on paper.He writes n integers a1, a2, ..., an. Each of those integers can be

#include <stdio.h>#include <string.h>char web[10000][75];int main(){ char command[10]; int i = 0,j; memset(web,'\0',sizeof(web)); while(1){ scanf("%s",command); if(command[0] == 'V'){ i++;

#include <cstdlib>#include <cstdio>#include <cstring>#include <iostream>#include <queue>#define INPUTusing namespace std;const int c0de4fun = 10;int x[4] = {1,-1,0,0};int y[4] = {0,0,1,-1};int maze[c0de4fun][c0de4fun];/*

 #include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <vector>#include <queue>#include <algorithm>//#define INPUTusing namespace std;/** Problem:POJ2442 - Sequence Begin

#include <stdio.h>#include <string.h>#include <math.h>int main(){ int num[500]; double in[500]; double ans; char op[500],io[500]; char c; int i,j,k,len; while(1){ memset(in,0,sizeof(in));

HDU-1542-Atlantishttp://acm.hdu.edu.cn/showproblem.php?pid=1542用線段樹求矩形面積的並，模仿別人的代碼寫的，還要好好研究啊#include<iostream>#include<cstring>#include<cstdlib>#include<cstdio>#include<algorithm>using namespace std;struct node{int

A. Ciel and Dancingtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputFox Ciel and her friends are in a dancing room. There are n boys and m girls here, and they never danced before. There will be

B. Hungry Sequencetime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputIahub and Iahubina went to a date at a luxury restaurant. Everything went fine until paying for the food. Instead of money, the

#include <stdio.h>#include <string.h>#include <stack>using namespace std;int T;stack <int> s[105];int main(){ int flag[105];//0 - up,1 - down int n,i,j; int pl,pr,p; int ans[105]; int outnum,out[105]; char

Hat’s WordsTime Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 7   Accepted Submission(s) : 4Font: Times New Roman | Verdana | GeorgiaFont Size: ← →Problem DescriptionA hat’s word is a word in the

#include <cstring>#include <cstdlib>#include <cstdio>#include <iostream>using namespace std;const int MAX_SIZE = 100010 * 2;const int INFINITE = 99999999;int num[MAX_SIZE];int sum[MAX_SIZE];int q[MAX_SIZE];int

Domino EffectTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6904 Accepted: 1754DescriptionDid you know that you can use domino bones for other things besides playing Dominoes? Take a number of dominoes and build a row by standing them on

I Hate ItTime Limit : 9000/3000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)Total Submission(s) : 6   Accepted Submission(s) : 2Problem

#include <iostream>#include <cstring>#include <cstdio>#include <cstdlib>/** Title: 【ZOJ3587】Marlon's String——白四爺×KMP 白瀨肆の演算法完全解讀KMP篇 KMP來襲第二彈首碼什麼的果然最討厭了！【1.0%達成！】 Problem: ZOJ3587 Marlon's String Knowledge Point :

#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>//#define INPUT/** Problem:1182 - 食物鏈，NOI2001 Begin Time:4th/Mar/2012 1:00 p.m. End Time:4th/Mar/2012 6:47 p.m. Cost Time:兩天多，看的別人的解題報告AC的

poj 1988num是統計該堆中cube的數目ans是統計該cube下面的cube的數目代碼如下：#include<iostream>using namespace std;#include<cstdio>const int maxn=50005;int pre[maxn],num[maxn],ans[maxn];void init(){ int i; for(i=0;i<maxn;i++){ pre[i]=i;

A. Magic Numberstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputA magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times.

題目用線段樹來求矩形面積並比照著其他人的代碼，總算理解了。先熟悉下，代碼如下：#include<iostream>#include<cstdio>#include<algorithm>using namespace std;#define maxn 220struct Node{ int l,r; int c; double len;}node[2000];struct Line{ double x,y1,y2; int

題目D 指將改點更新為0Q 查詢改點處左右兩端的連續區間R 將前一個D的點回複成1典型的區間合并的題目，代碼：#include<iostream>#include<cstdio>using namespace std;#define maxn 50005#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1struct Tree{ int l,r; int