其他

    這題是明顯的貪心，首先在讀取資料時把敵人按有刀和無刀分成兩類，剛開始有兩種貪心決策，只殺無刀的人和殺掉所有有刀的人，然後比較這兩種決策，選出最優解。   

找出數組中重複的數前序        最近一直在看v_JULY_v的專欄，從中學到了很多關於演算法方面的知識，也受到了很大的啟發。我相信喜歡演算法的朋友，看過他的博文之後也會有這種想法。前段時間參加了一些面試，從他的專欄裡學到的演算法給予了我不小的協助，這讓我在面試的時候輕鬆了不少。他的博文我將會繼續關注和學習。        好了，我們言歸正傳。關於找出數組中重複數的這個問題，網上已經有很多文章來闡述。不過我認為各位對於這個問題考慮的情況不是很全面，於是自己就想把這個問題完善一下。     

         首先來說一下此題的動歸思想。可以類比一下我們手放的過程，剛開始時棋盤為空白，所以我們就從第一行開始鋪，當第一行鋪滿後，再從第二行開始鋪（注意此時第二行有些方格可能已經被佔了），然後一直進去……           

在MSDN中這樣描述：If you wish to create a modeless dialog, call Createin the constructor of your dialog class.When you implement a modeless dialog box, always override the OnCancel member function and callDestroyWindow from within it. Don't call the base

Drainage DitchesProblem DescriptionEvery time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has

    題意：在一個串的後面添加若干字元，使其成為一個迴圈串。/*利用next數組求迴圈節結論：若且唯若len%(len-next[len-1])==0時,s[next[len-1]~len-1]為最小迴圈節*/#include<stdio.h>const int N=100005;char s[N];int next[N],len;void getnext(){ int i,j; next[0]=0; for(i=1,j=0;s[i];i++) {

大話編譯原理---上篇前序       記得第一次上編譯原理這門課時，老師曾慷慨激昂的說：“學好編譯原理能讓你們享用一生，你們要好好學啊”。不過學完編譯原理也有一段時間了，平時也找一些編譯原理方面的資料學習，卻始終感受不到學習編譯原理的效用。究其原因還是自己學的太單薄了，畢竟像這種“內功”可不是一天兩天就能練出來的！只是感覺這門課挺有意思，權當一門興趣來學了。本文       本文將介紹六大節的內容，其中每個部分又分為若干小節，如描述：第一節 編譯器概覽      

Optimal MilkingTime Limit: 2000MS Memory Limit: 30000KTotal Submissions: 9624 Accepted: 3474Case Time Limit: 1000MSDescriptionFJ has moved his K (1 <= K <= 30) milking machines out into the cow pastures among the C (1 <= C <= 200) cows.

Holedox MovingTime Limit: 5000MSMemory Limit: 65536KTotal Submissions: 11975Accepted: 2873DescriptionDuring winter, the most hungry and severe time, Holedox sleeps in its lair. When spring comes, Holedox wakes up, moves to the exit of its lair,

    思路：轉化為母函數求種數，x的指數表示價值，係數表示種數。則對於第一個範例，可表示為(x+1)*(x^2+1)*(x^3+1),展開為x^6+x^5+x^4+2x^3+x^2+x+1,所以種數為1+1+1+2+1+1=7。#include<iostream>#include<cstdio>using namespace std;int num[26];int a[51],b[51];int main(){ int t,i,j,k,ans;

Power NetworkTime Limit: 2000MSMemory Limit: 32768KDescriptionA power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied with an amount s(u) >= 0 of power, may

Basic Wall MazeTime Limit: 1000MSMemory Limit: 65536KTotal Submissions: 2548Accepted: 1141Special JudgeDescriptionIn this problem you have to solve a very simple maze consisting of:a 6 by 6 grid of unit squares3 walls of length between 1 and 6 which

Frame StackingTime Limit: 1000MSMemory Limit: 10000KTotal Submissions: 3636Accepted: 1204DescriptionConsider the following 5 picture frames placed on an 9 x 8 array......... ........ ........ ........ .CCC....EEEEEE.. ........ ........ ..BBBB.. .C.C.

A Plug for UNIXTime Limit: 1000MSMemory Limit: 65536KTotal Submissions: 12438Accepted: 4136DescriptionYou are in charge of setting up the press room for the inaugural meeting of the United Nations Internet eXecutive (UNIX), which has an

Following OrdersTime Limit: 1000MSMemory Limit: 10000KTotal Submissions: 3040Accepted: 1160DescriptionOrder is an important concept in mathematics and in computer science. For example, Zorn's Lemma states: ``a partially ordered set in which every

Ombrophobic BovinesTime Limit: 1000MSMemory Limit: 65536KTotal Submissions: 11558Accepted: 2566DescriptionFJ's cows really hate getting wet so much that the mere thought of getting caught in the rain makes them shake in their hooves. They have

#include<iostream>#include<cstdio>using namespace std;int ans[10000000];//存放n！的結果int main(){ int n,i,j,k,len; while(~scanf("%d",&n)) { ans[1]=1; len=1; for(i=2;i<=n;i++) { for(j=1;j<=len;j++)//邊乘邊進位 { ans[j]=ans[j]*i+

Labeling BallsTime Limit: 1000MSMemory Limit: 65536KTotal Submissions: 8814Accepted: 2387DescriptionWindy has N balls of distinct weights from 1 unit toN units. Now he tries to label them with 1 to N in such a way that:No two balls share the same

深入解析最短路徑演算法本文  第一節 問題的提出及解決方案       所謂最短路徑問題，可以說有兩種情況來描述。       描述一：在圖論中，指的是尋找圖中兩個節點之間的最短距離。如       描述二：在現實生活中，指的是找到從一個地方到另一個地方的最近距離。如       上述兩種情況的本質是一樣的，即求一個點到另一個點的最短路徑。好了，問題已經提出來了，那怎麼解決呢？解決該問題的方法還是比較多的，不過由於各個路徑演算法所對應的問題條件不同，我們可根據不同的情況，選擇不同的路徑演算法。