cf194 div2 A. Candy Bags

A. Candy Bagstime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputGerald has n younger brothers and their number happens to be even. One day he bought n2 candy bags. One bag has one candy, one bag has

cf192 div2 A. Cakeminator

A. Cakeminatortime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given a rectangular cake, represented as an r × c grid. Each cell either has an evil strawberry, or is empty. For example, a 3 

cf 194 div2 BEight Point Sets

B. Eight Point Setstime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputGerald is very particular to eight point sets. He thinks that any decent eight point set must consist of all pairwise

HDU-1431-素數迴文

HDU-1431-素數迴文http://acm.hdu.edu.cn/showproblem.php?pid=1431看了題目Statistic的代碼長度就估計要打表先找出所有的迴文素數#include<stdio.h>#include<string.h>#include<stdlib.h>#define N 100000005char a[N];int num[100000],t;int huiwen(int x) {int

zoj 2913 BFS

//zoj 2913 BFS// 從每條線路上的每個地區出發進行BFS遍曆,// 統計每條線路上每個地區到其他地區最短距離中的最大值,求出所有最大值中的最小值。#include<stdio.h>#include<queue>#include<string.h>using namespace std;#define MAX 10000#define INF 100000int nz,nr,cur;int mz[MAX];int

cf 193 div2 A. Down the Hatch!

A. Down the Hatch!time limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputEverybody knows that the Berland citizens are keen on health, especially students. Berland students are so tough that all they

cf 192 div2 B. Road Construction

B. Road Constructiontime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputA country has n cities. Initially, there is no road in the country. One day, the king decides to construct some roads

【POJ2488】 A knight’s Journey 解題報告 測試資料+代碼+思路

 #include <cstdio>#include <cstdlib>#include <cstring>#include <queue>#include <iostream>//#define INPUT//#define DBG/** Problem:POJ2488 Begin Time:8th/Mar/2012 1:30 p.m. End Time:2012-03-08 21:08:53 Test

【poj1164】The Castle,解題報告+思路+代碼+資料

#include <iostream>#include <cstring>#include <cstdio>#include <cstdlib>#include <stack>#include <queue>#define INPUTusing namespace std;/** Problem: POJ1164 Begin Time:7th/3/2012 10.00 p.m. End Time:2012-

【POJ3740】Easy Finding,解題報告+思路+代碼

#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>//#define INPUT/** Problem: poj3740 - Easy Finding Begin Time:8:30 p.m. 20th/Mar/2012 End Time: 8:42 p.m. 21st/Mar/2012 Last Time: Maybe

【HDU4313】 – Matrix – 樹狀DP Version 思路+解題報告+AC代碼【0.4%達成】

#include <cstring>#include <cstdlib>#include <cstdio>#include <algorithm>#include <iostream>using namespace std;/** Problem: HDU4313 - Matrix - DP Version【0.4%達成】 Copyright : 歸我們學校集訓隊和本人所有,未經同意可以轉載。

快速冪乘

快速冪乘用的是二分的思想a^b%c,當b比較大時可將其分解當b為偶數時,a^b%c=(a^(b/2)*a^(b/2))%c;當b為奇數時,a^b%c=(a^(b/2)*a^(b/2)*a)%cAOJ-569-乘的更快http://icpc.ahu.edu.cn/OJ/Problem.aspx?id=569赤裸裸的快速冪乘#include<stdio.h>#include<string.h>#include<stdlib.h>#define Mod 99991_

8數位問題

8數位問題,即在一個3×3的矩陣中有8個數(1至8)和一個空格,從一個狀態轉換到另一個狀態,每次只能移動與空格相鄰的一個數字到空格當中AOJ-417-8數位http://icpc.ahu.edu.cn/OJ/Problem.aspx?id=417這題是求轉化的最少步數,可用BFS解決,共有9!=362880種情況,關鍵是如何標記已經訪問過的狀態,保證每次搜尋得到的狀態都是最小的步數,這裡可將字串轉化成對應的整數來處理,可用康托展開來節省儲存空間康托展開: X=an*(n-1)!+an-1*(n-

關於 pow 浮點精度丟失

/* 寢室學弟問我一個很簡單的問題,就是把一個數逆序後輸出的字串轉換為數值,學弟用了pow這個很常見的函數,代碼邏輯沒有問題,但是不理解為什麼答案是錯的,而且只差1或者2;在小蒙的協助下,才開始意識到pow 精度丟失的問題.一查 cplusplus.com 才發現 如下問題: double pow ( double base, double exponent );long double pow ( long double base, long double

解決tomcat startup.bat啟動時一閃即過問題

tomcat可以在myeclipse下啟動使用,但單獨通過startup.bat啟動雙擊後視窗一閃即逝。主要原因為tomcat在啟動的過程中需要尋找jre的安裝路徑,因此在啟動tomcat前,需要首先安裝JDK,然後配置好環境變數,有很多時候是因為環境變數配置的不正確而導致tomcat啟動不起來。下面以我的安裝路徑為例。環境變數的配置如下:CATALINA_BASE D:\apache-tomcat-7.0.6    --tomcat的安裝根路徑CATALINA_HOME

高精度計算

// 高精度加法#include<iostream>#include<cstring>using namespace std;int* strtoint(char *str){ int i,len=strlen(str); int* a=new int[(len+1)*sizeof(int)]; for(i=0;i<len;i++) a[i]=(int)str[len-i-1]-48; return a;}char*

把一切推翻,重新再來

渾渾噩噩恍恍惚惚走到了大二下學期,這兩年總是感覺什麼都急著速成,卻什麼都沒有做成。所以現在,一拳揍倒以前的自己,重新再來吧。 關鍵詞:WEB搜尋, WEB挖掘, 網路爬蟲,

【POJ1011 Sticks】解題報告+思路+代碼

#include <cstring>#include <cstdio>#include <cstdlib>#include <algorithm>#include <iostream>//#define INPUTusing namespace std;/** Problem : poj1011 - Sticks Begin Time : 13:00 p.m. 15th/mar/2012 End Time : 15:1

HDOJ 1069 Monkey and Banana (動態規劃)

/*結題報告:DP第一題, 不怎麼會,看了下discuss才自己敲的。心得: 其實蠻簡單的,只有sort()方法是關鍵,length 和 weight 都必須比下一個大,用area排序比較合理 */#include <iostream>#include <algorithm>#include <iomanip>using namespace std;#define MAX 35struct block{int x, y, z, f;}b[120];void

【HDU2780 – Su-Su-Sudoku】 解題報告+思路+代碼,差點一次AC T^T

#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>#include <cstdlib>////#define INPUT///#define DBG/** Problem: HDU2780 - Su-Su-Sudoku Begin Time: 19th/Mar/2012 17:52 End Time:

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