275. To xor or not to xor The sequence of non-negative integers A1, A2, ..., AN is given. You are to find some subsequence Ai 1, Ai 2, ..., Ai k (1 <= i 1 < i 2 < ... < i k<= N) such, that Ai 1 XOR Ai 2 XOR ... XOR Ai k has a maximum valu…
#include<stdio.h> int max(int a,int b,int c); int main() { int a,b,c; while(scanf("%d %d %d",&a,&b,&c)!=EOF){ printf("%d\n",max(a,b,c));} ; } int max(int a,int b,int c) { int m; m=a; if(b>m) m=b; if(c>m) m=c; re…
//求n个数中的最小k个数 public static void TestMin(int k, int n) { Random rd = new Random(); int[] myArray = new int[n]; int[] newArray = new int[k]; for (int i = 0; i < n; i++) { // rand…
[本文出自天外归云的博客园] 题1:求m以内的素数(m>2) def find_all_primes_in(m): def prime(num): for i in range(2, num): if divmod(num, i)[1] == 0: return False return True print([i for i in range(2, m + 1) if prime(i)]) if __name__ == '__main__': find_all_primes_in(100) 我…
//求两个数中不同的位的个数 #include <stdio.h> int count_different(int a, int b) { int count = 0; int c = a^b; //a,b中不同的位即为1 while (c) { count++; c = c&(c - 1); //把c中最后一个1去掉 } return count; } int main() { printf("%d\n", count_different(3,8)); //3 p…
函数重载重载的思想很简单:编译器允许你用同一名字定义多个函数或过程,只要它们所带的参数不同.实际上,编译器是通过检测参数来确定需要调用的例程.下面是从VCL 的数学单元(Math Unit)中摘录的一系列函数: function Min (A,B: Integer): Integer; overload; function Min (A,B: Int64): Int64; overload; function Min (A,B: Single): Single; overload; functi…
package wac.wev.as;//新建一个方法在求最大值import java.util.Scanner; public class MaxLian {public static void main(String[] args){//键盘录入以及导包Scanner sc= new Scanner(System.in);//数据接收System.out.println("请输入第一个数据:");int a = sc.nextInt();System.out.println(&qu…
