//判定数组是否有序 //总的程序代码如下: using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Threading.Tasks; namespace ConsoleApplication5 { class Program { public static void Main(string[] args) { ; bool result; Console
A. Olympiad time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output The recent All-Berland Olympiad in Informatics featured n participants with each scoring a certain amount of points. As the head
package com.wh.array; public class Lottery { public static void main(String[] args) { int[] num=new int[11]; //11个球 for (int i = 0; i < num.length; i++) { num[i]=i+1; } //打乱排序 for (int i = 0; i < num.length; i++) { int ran=(int)(Math.random()*11); i
//描述:利用可变参数列表统计一组数的平均值 #include <stdarg.h> #include <stdio.h> float average(int num, ...);//函数原型:即声明 float average2(int num, ...);//num个数 void add(int num, int x, int y, int z); int main(void){ int a=10; int b=20; printf("a地址:%p b地址:%p\n&
N=int(raw_input('input the number N=')) number=[] while(N): m=int(raw_input('input..\n')) number.append(m) N -=1 print number for i in range (0,len(number),1): for j in range(0,i,1): if number[i]<number[j]: t=number[j] number[j]=number[i] number[i]=t
问题描述:找出一组数字出现次数最多的数,如果有多个这样的数,输出其中最小的一个. 算法:sort排序,遍历数组,每遍历一个数,查出它已经出现的次数. 代码: #include<bits/stdc++.h> using namespace std; int main() { int n; cin>>n; int a[1005]; for(int i=0;i<n;i++) { cin>>a[i]; } sort(a,a+n); int k=0,flag=0,max=0
前m大的数 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 19800 Accepted Submission(s): 6781 Problem Description 还记得Gardon给小希布置的那个作业么?(上次比赛的1005)其实小希已经找回了原来的那张数表,现在她想确认一下她的答案是否正确,但是整个的答案是很庞大的表,小希
Priest John's Busiest Day Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 8170 Accepted: 2784 Special Judge Description John is the only priest in his town. September 1st is the John's busiest day in a year because there is an old le
