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ZOJ 3235 Prototype
PrototypeTime Limit: 1 Second Memory Limit: 32768 KBPrototype is a 3D game which allow you to control a person named Alex with much super ability t...
分类:其他好文   时间:2014-07-27 11:06:42    阅读次数:328
import the library to Android Studio
To import the library to Android Studio, there is two methods that can (or cannot) work. The first one worked for me, but when I tried the second, it ...
分类:移动开发   时间:2014-07-24 22:42:13    阅读次数:284
从第一个字符串中删除第二个字符串中出现过的所有字符
// 从第一个字符串中删除第二个字符串中出现过的所有字符#include char* remove_second_from_first( char *first, char *second ){ if( first == NULL || second == NULL ) { ...
分类:其他好文   时间:2014-07-24 09:58:23    阅读次数:163
JavaScript Patterns 7.1 Singleton
The idea of the singleton pattern is to have only one instance of a specific class. This means that the second time you use the same class to create a...
分类:编程语言   时间:2014-07-24 05:01:38    阅读次数:422
Codeforces Round #257 (Div. 2) E题:Jzzhu and Apples 模拟
E. Jzzhu and Apples time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Jzzhu has picked n apples from his big a...
分类:移动开发   时间:2014-07-23 22:37:47    阅读次数:387
javascript将毫秒转换成hh:mm:ss的形式
function formatMilliseconds(value) { var second = parseInt(value) / 1000; // second var minute = 0; // minute var ho...
分类:编程语言   时间:2014-07-23 20:40:55    阅读次数:195
MySQL时区处理
开发人员执行如下SQL root@localhost{wm_ztcj}?>select??timestampdiff(second,‘1970-1-1‘,‘2014-07-23?9:18:40‘)?as?timestamp; +------------+ |?timestamp??| +------------+ |?1406107120?| +----...
分类:数据库   时间:2014-07-23 13:57:06    阅读次数:483
pair的用法
初始化: std::pair p; //initialize p.first and p.second with zero std::pair p(42, "hello"); make_pair(42, "hello"); // no need for the var name, it's r...
分类:其他好文   时间:2014-07-22 22:47:13    阅读次数:209
(转)压力测试的轻量级具体做法
一:压力测试中需要掌握的几个基本概念1:吞吐率(Requests per second)服务器并发处理能力的量化描述,单位是reqs/s,指的是某个并发用户数下单位时间内处理的请求数。某个并发用户数下单位时间内能处理的最大请求数,称之为最大吞吐率。记住:吞吐率是基于并发用户数的。这句话代表了两个含义...
分类:其他好文   时间:2014-07-22 00:37:34    阅读次数:215
从键盘上输入两个数,按小大的顺序输出
#include #include int main() {    int num1,num2;    int *num1_p=&num1,*num2_p=&num2,*pointer;    printf("Input the first number:");    scanf("%d",num1_p);    printf("Input the second num...
分类:其他好文   时间:2014-07-21 11:14:14    阅读次数:180
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