代码: using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Threading.Tasks; namespace 委托_例子 { static class Program { // 定义委托(Double类型) delegate double Integand(double x); //定义方法 static double Method1(double
The area Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1637 Accepted Submission(s): 1298 Problem Description Ignatius bought a land last week, but he didn't know the area of the land because t
LOOPS Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 125536/65536 K (Java/Others) Total Submission(s): 1864 Accepted Submission(s): 732 Problem Description Akemi Homura is a Mahou Shoujo (Puella Magi/Magical Girl). Homura wants to help h
关于欧几里得算法求最大公约数算法, 代码如下: int gcd( int a , int b ) { if( b == 0 ) return a ; else gcd( b , a % b ) ; } 证明: 对于a,b,有a = kb + r (a , k , b , r 均为整数),其中r = a mod b . 令d为a和b的一个公约数,则d|a,d|b(即a.b都被d整除), 那么 r =a - kb ,两边同时除以d 得 r/d = a/d - kb/d = m (m为整数,因为r也
素数:一个数只能被1和它本身整除的数.2是最小的素数 #include <iostream> using namespace std; #define NUM 100 ]; int main() { //筛选法求素数 //假设所有的素数都是素数,标志位设为1 ; i <= NUM ; i++){ isPrime[i] = ; } // 首先去除当前数的倍数.例如当前数为2,那么去除4,6,8等等 ; i <= NUM ; i++){ if(isPrime[i]){ //将相应的标志
快排法求第k大,复杂度为O(n) import com.sun.media.sound.SoftTuning; import java.util.Arrays; import java.util.Random; public class Main { int[] generate(int n) { Random random = new Random(); int[] a = new int[n]; for (int i = 0; i < a.length; i++) { a[i] = rand
原文地址: http://blog.csdn.net/qustmeng/article/details/52186378?locationNum=4&fps=1 import java.util.LinkedList; import java.util.List; public class Demo { /** * 二次指数平滑法求预测值 * @param list 基础数据集合 * @param year 未来第几期 * @param modu
How far away ? Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 14685 Accepted Submission(s): 5554 Problem Description There are n houses in the village and some bidirectional roads connecting