下次不要被长题目吓到,其实不一定难. 先看输入输出,再揣测题意. 原文: 2548.   Celebrity jeopardy Time Limit: 1.0 Seconds   Memory Limit: 65536KTotal Runs: 1306   Accepted Runs: 898 It's hard to construct a problem that's so easy that everyone will get it, yet still difficult enough…

1.Link: http://poj.org/problem?id=3062 2.Content: Celebrity jeopardy Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 15015   Accepted: 8471 Description It's hard to construct a problem that's so easy that everyone will get it, yet still…

题目描写叙述 Problem C: Celebrity Split Jack and Jill have decided to separate and divide their property equally. Each of their N mansions has a value between 1,000,000 and 40,000,000 dollars. Jack will receive some of the mansions; Jill will receive some…

It's hard to construct a problem that's so easy that everyone will get it, yet still difficult enough to beworthy of some respect. Usually, we err on one side or the other. How simple can a problem really be? Here, as in Celebrity Jepoardy, questions…

注意代码中: result1 << " to " << result2 << ", PLAYER 1 WINS."<< endl; 和 result1 << " to " << result2 << ", PLAYER 1 WINS. "<< endl; 虽然只在WINS后只差一个空格,但会导致PE. 原题: 2101.   Bul…

原题: 2857.   Digit Sorting Time Limit: 1.0 Seconds   Memory Limit: 65536KTotal Runs: 3234   Accepted Runs: 1704 Several players play a game. Each player chooses a certain number, writes it down (in decimal notation, without leading zeroes) and sorts t…

最重要的是找规律. 下面是引用 http://blog.sina.com.cn/s/blog_4dc813b20100snyv.html 的讲解: 做这题时,千万不要被那个图给吓着了,其实这题就是道简单的数学题. 首先看当m或n中有一个为2的情况,显然,只需要算周长就OK了.即(m+n-)*,考虑到至少其中一个为2,所以答案为2 *m或2*n,亦即m*n.注意这里保证了其中一个数位偶数. 当m,n≥3时,考虑至少其中一个为偶数的情况,显然,这种情况很简单,可以得出,结果为m*n,又可以和上面这种…

注意数据范围,十位数以上就可以考虑long long 了,断点调试也十分重要. 原题: 1065.   Factorial Time Limit: 1.0 Seconds   Memory Limit: 65536KTotal Runs: 6067   Accepted Runs: 2679 The most important part of a GSM network is so called Base Transceiver Station (BTS). These transceiver…

注: 1. 对于double计算,一定要小心,必要时把与double计算相关的所有都变成double型. 2. for (int i = 0; i < N; i++)         //N 不可写为N - 1,否则当N为1时无法进行: 原题: 1100.   Pi Time Limit: 1.0 Seconds   Memory Limit: 65536KTotal Runs: 5683   Accepted Runs: 2317 Professor Robert A. J. Matthews…

注意: for (int i = 1; i <= aaa.length(); i++) 其中是“ i <= ",注意等号. 原题: 2520.   Quicksum Time Limit: 0.5 Seconds   Memory Limit: 65536KTotal Runs: 2964   Accepted Runs: 1970 A checksum is an algorithm that scans a packet of data and returns a single…