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早7:30在东直门长途客车站集结,坐936到云蒙山公园沿111国道步行2km到后山铺,开始穿越之旅
没有GPS设备,但可以利用GPS数据实现GPS数据程序:输入 点a,N40.60247 E116.78447和点b,N40.60525 E116.77914的坐标信息输出 b对于a的角度,例如北偏西25度;以及a,b间长度网上有人说如果不考虑地球的扁平率误差会很大不过找到了这个网址:http://www.movable-type.co.uk/scripts/LatLong.html用这个网页测了02 N40.60525 E116.77914 途中的岔路,走左边03 N40.60556 E116.77136 308m 水坝,前方是水泉峪(废墟)02,02点间距离=0.6577 km用Google Earth测的距离是0.66 km看来这个网页还是可以用的,我已经当下来存到手机里了这样依靠GPS数据,就可以确定下一个点在什么方位,在多少米距离上,迷路以及走弯路的可能就少了
Archive for September, 2006
2006年10.3云蒙山穿越Trek准备清单Checklist
编写一个函数,作用是把一个char组成的字符串循环右移n个
例如,”abcdef”循环右移2个就变为”efabcd”
popo实现了一个,不过需要O(n)的额外空间程序:void LoopMove(char * pStr,int steps){ int len = strlen(pStr); char *s = new char[len+1]; int count = 0; char *temp = pStr; while (*pStr) { s[(count+steps)%len] = *pStr; pStr++; count++; } s[len] = ‘\0′; strcpy(temp,s); delete []s;}
最近想出了个空间为O(1),时间为O(n)的算法就是只通过数组指针移动和交换来实现
int LoopMove2(char *p, int i, int t, int s, int m){ int n = strlen(p); int [...]
Google 面试第一题解法
Solve this cryptic equation, realizing of course that values for M and E could be interchanged. No leading zeros are allowed.
WWWDOT - GOOGLE = DOTCOM
思路:穷举法,式子中共有9个不同的字母,数字0-9有10个,因此就从10个数字中任取9个,排成一列,这样分别对应了式子中的字母。然后判断当前取出的这9个数,是否满足条件。本程序中用check2 函数判断,另外需要考虑WE是可以互换,但不可以相同
答案:只有两组解777589 - 188103 = 589486777589 - 188106 = 589483
程序:Perm1.javapackage org.chao.perm;
public class Perm1 { public static void main(String[] args) { [...]
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