Google Page 挑戰 MSN Space 今天,我的google page 帳號申請成功了,google 好像總是這樣,走Gmail之路,不會讓所有人想什麼時候註冊就什麼時候註冊,給人的感覺得到太容易,放棄的也太容易,很有意思哦。 剛剛上去看了下,還沒有時間去做,感覺和MSN Space 差不多,不過沒有msn做得好。不過我們需要等待,等待google給我們的驚喜,因為google太有創意了。
linux 中很多系統調用的api 在函式宣告的地方都有一個: asmlinkage http://kernelnewbies.org/FAQ/asmlinkage 中說:The asmlinkage tag is one other thing that we should observe about this simple function. This is a #define for some gcc magic that tells the compiler that the
Floyd-Warshall 演算法用來找出每對點之間的最短距離。它需要用鄰接矩陣來儲存邊,這個演算法通過考慮最佳子路徑來得到最佳路徑。 注意單獨一條邊的路徑也不一定是最佳路徑。 從任意一條單邊路徑開始。所有兩點之間的距離是邊的權,或者無窮大,如果兩點之間沒有邊相連。 對於每一對頂點 u 和 v,看看是否存在一個頂點 w 使得從 u 到 w 再到 v 比己知的路徑更短。如果是更新它。 不可思議的是,只要按排適當,就能得到結果 // dist(i,j)
Here is my bash alias. The argument is the <branch name> to export.An example: cd /path/to/foo/ _gittar masterThe output is (time in ISO 8601 / UTC): ../foot-YYYYMMDD.HHMM.tar.gzfunction _gittar (){ local name=$(pwd) name=${name#
The oneway keyword means that if that call results in an IPC (i.e. the caller and callee are in different processes) then the calling process will not wait for the called process to handle the IPC. If it does not result in an IPC (i.e. they're both
DP 有兩種類型,一種是:Top-Down 一種是:Bottom-Up 。 DP 的典型的應用有幾種: 1、Coin change : is the problem of finding the number of ways of making changes for a particular amount of cents,n, using a given set of denominations d[1...m]. The problem is defined follows:
You have a five-quart jug, a three-quart jug, and an unlimited supply of water (but no measuring cups). How would you come up with exactly four quarts of water? Note that the jugs are oddly shaped, such that filling up exactly "half" of the jug
據說是HuaWei 面試試題,不知道真假了。#include <stdio.h>int check(int m){ int n =m ,i,d = 0; while(n) { d = d * 10 + n %10; n /= 10 ; } return (d == m);}int checkstr(char * str){ int len,