本渣清明节 闲里偷忙 做了一下codejam 水平不出意外的在投稿之后一落千丈 后两题的hidden test竟然都挂了 A. Foregone Solution 水题,稍微判断一下特殊情况(比如1000, 5000这种)就好了 #include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <climits> #include <cstr…

Google Code Jam Africa 2010 Qualification Round Problem B. Reverse Words https://code.google.com/codejam/contest/351101/dashboard#s=p1 Problem Given a list of space separated words, reverse the order of the words. Each line of text contains L letters…

Google Code Jam Qualification Round Africa 2010 Problem A. Store Credit https://code.google.com/codejam/contest/351101/dashboard#s=p0 Problem You receive a credit C at a local store and would like to buy two items. You first walk through the store an…

Facebook Hacker Cup 2014 Qualification Round比赛Square Detector题的解题报告.单击这里打开题目链接(国内访问需要那个,你懂的). 原题如下: Square Detector Problem Description You want to write an image detection system that is able to recognize different geometric shapes. In the first ver…

题目地址:http://blog.csdn.net/shiyuankongbu/article/details/10004443 /* 题意:在i前面找回文子串,在i后面找回文子串相互配对,问有几对 DP:很巧妙的从i出发向两头扩展判断是否相同来找回文串 dpr[i] 代表记录从0到i间的回文子串的个数,dpl[i] 代表记录i之后的回文子串个数 两两相乘就行了 详细解释:http://blog.csdn.net/shiyuankongbu/article/details/10004443 */…

2014 Qualification Round Solutions 2013年11月25日下午 1:34 ...最简单的一题又有bug...自以为是真是很厉害! 1. Square Detector (20 Points) When facing a problem in a programming contest there are three main things to consider when planning your solution. In order of importanc…

Google Code Jam Qualification Round Africa 2010 的第一题,很简单. Problem You receive a credit C at a local store and would like to buy two items. You first walk through the store and create a list L of all available items. From this list you would like to b…

Problem B. Infinite House of Pancakes Problem's Link:   https://code.google.com/codejam/contest/6224486/dashboard#s=p1 Mean: 有无限多个盘子,其中有n个盘子里面放有饼,每分钟你可以选择两种操作中的一种: 1.n个盘子里面的饼同时减少1: 2.选择一个盘子里面的饼,分到其他盘子里面去: 目标是让盘子里的饼在最少的分钟数内吃完,问最少的分钟数. analyse: 可以分析出,先…

Problem A. Standing Ovation Problem's Link:   https://code.google.com/codejam/contest/6224486/dashboard#s=p0 Mean: 题目说的是有许多观众,每个观众有一定的羞涩值,只有现场站起来鼓掌的人数达到该值才会站起来鼓掌,问最少添加多少羞涩值任意的人,才能使所有人都站起来鼓掌. analyse: 贪心模拟一下,从前往后扫一遍就行. Time complexity: O(n) Source cod…

本题的 Large dataset 本人尚未解决. https://code.google.com/codejam/contest/90101/dashboard#s=p2 Problem So you've registered. We sent you a welcoming email, to welcome you to code jam. But it's possible that you still don't feel welcomed to code jam. That's w…