其他

Problem Description假設有x1個字母A， x2個字母B,..... x26個字母Z，同時假設字母A的價值為1，字母B的價值為2,.....

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=709The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large rectangular region of land at a time, and creates a grid

題目大意：有一個h*w的公告榜，可以依次在上面添加資訊。每個資訊的長度為x，高為1. 優先在最上面加入，如果空間足夠的話，然後優先放在最左面。統計每條公告最終的位置，即它所在的行數。這裡是線段樹來儲存 當前區間(i,j)的所有位置，剩餘的最大空間。 初始化即為w，公告榜的寬。Problem DescriptionAt the entrance to the university, there is a huge rectangular billboard of size h*w (h is

成段更新,需要用到延遲標記(或者說懶惰標記),簡單來說就是每次更新的時候不要更新到底,用延遲標記使得更新延遲到下次需要更新or詢問到的時候.此處建議在紙上類比一遍。Problem DescriptionIn the game of DotA, Pudge’s meat hook is actually the most horrible thing for most of the heroes. The hook is made up of several consecutive

可參考這個題http://blog.csdn.net/gaotong2055/article/details/9300141DescriptionYou have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The

此題為基礎的線段樹。不用建樹，可直接用數組類比。Problem

FatMouse's SpeedTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6929    Accepted Submission(s): 3028Special JudgeProblem DescriptionFatMouse believes that the fatter a mouse is, the faster it

當n個編號元素放在n個編號位置，元素編號與位置編號各不對應的方法數用M(n）表示，那麼M(n-1）就表示n-1個編號元素放在n-1個編號位置，各不對應的方法數，其它類推.第一步，把第n個元素放在一個位置，比如位置k，一共有n-1種方法；第二步，放編號為k的元素，這時有兩種情況⑴把它放到位置n，那麼，對於剩下的n-1個元素，由於第k個元素放到了位置n，剩下n-2個元素就有M(n-2）種方法；⑵第k個元素不把它放到位置n，這時，對於這n-1個元素，有M(n-1）種方法；綜上得到M(n)=(n-1）[

首先看多次相加String 迴圈10000次，後兩者是100000，累加次數是String的10倍long start = System.currentTimeMillis();String result = "";for (int i = 0; i < 10000; i++) {String s1 = "hello";String s2 = "world";int s3 = 1000;result += s1;result += s2;result +=

Problem DescriptionMarsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbles. This would be easy if all the marbles had the same value, because then they could

例如：6 有如下11種劃分則p(6)=116;5+1;4+2, 4+1+1;3+3, 3+2+1, 3+1+1+1;2+2+2, 2+2+1+1, 2+1+1+1+1;1+1+1+1+1+1;在正整數n所有劃分中，將最大加數n1不大於m的劃分個數記作q(n, m).我們可以建立如下遞迴關係:(1) q(n, 1) = 1, n>=1最大加數不大於1,則只有一種劃分，n=1+1+1+...+1(2) q(n, m) = q(n, n), m>=n最大加數n1實際上不能大於n.因此q(1

題目描述：給你一串路徑，譬如：a\b\ca\d\eb\cstd\你把這些路徑中蘊含的目錄結構給畫出來，子目錄直接列在父目錄下面，並比父目錄向右縮一格，就像這樣：a  b    c  d      eb  cstd同一級的需要按字母順序排列，不能亂。輸入：   

以二階矩陣為例。典型的斐波那契數列，就可以使用矩陣相乘來求解。如果不考慮乘法的話，複雜度應該是lgn.下面只給出計算矩陣快速冪的方法#include <iostream>#define L 10000using namespace std;class MyMatrix{public:long long a00,a01,a10,a11;MyMatrix();MyMatrix(long long _a00, long long _a01, long long _a10, long

Problem Description把一個偶數拆成兩個不同素數的和，有幾種拆法呢？ Input輸入包含一些正的偶數，其值不會超過10000，個數不會超過500，若遇0，則結束。 Output對應每個偶數，輸出其拆成不同素數的個數，每個結果佔一行。 Sample Input30260 Sample Output32#include <iostream>#include <string.h>using namespace std;int prim[10001];int

為保證裝置的正常運行，必須配備一定數量的裝置維修人員，現有同類裝置300台，且各台裝置工作相互獨立，任一時刻發生故障的機率都是0.01，假設一台裝置的故障由一人進行修理，問至少應配備多少名修理人員，才能保證裝置發生故障後能得到及時修理的機率不小於0.99？當N大，p小時，可以用此公式，是近似的二項分布計算，相對的計算量會少一些。當然，對於此題，精確計算應該是用二項分布，但是計算C（n,k）過大，會溢出。下面的程式得出至少需要8名人員。#include

DescriptionBefore bridges were common, ferries were used to transport cars across rivers. River ferries, unlike their larger cousins, run on a guide line and are powered by the river's current. Cars drive onto the ferry from one end, the ferry

Problem Description"Well, it seems the first problem is too easy. I will let you know how foolish you are later." feng5166 says."The second problem is, given an positive integer N, we define an equation like

DescriptionFarmer John has a pig farm near town A. He wants to visit his friend living in town B. During this journey he will visit n small villages so he decided to earn some money. He tooks n pigs and plans to sell one pig in each village he

Problem DescriptionWe all know that Bin-Laden is a notorious terrorist, and he has disappeared for a long time. But recently, it is reported that he hides in Hang Zhou of China! “Oh, God! How terrible! ”Don’t be so afraid, guys. Although he hides in

#include <iostream>using namespace std;int x[20]; //解向量int sum; //可行方案個數int n;//在第t列是否可放置bool place(int t){int i;for(i=1; i<t; i++)//如果 行-行 == 列-列 或者 在同一列。if(abs(t-i) == abs(x[t]-x[i]) || x[i]==x[t])return false;return true;}//回溯，即遞迴調用void