var n = readInt();
var arr=new Array(n);
for(i=0;i<n;i++){
arr[i]=readInt();
}
var compare = function(x,y){
if(x<y){
return -1;
}else if(x>y){
return 1;
}else{
return 0;
}
};
arr.sort(compare);
if(arr.length%2==0){
console.log((arr[arr.length/2-1]+arr[arr.length/2])/2);
}else{
console.log(arr[(arr.length-1)/2]);
}
第一题，完全部图，c++ ac答案
#include <iostream>
#include <vector>
#include <set>
#include <algorithm>
using namespace std;
int findunlink(vector<int>& table,int n)
{
for(int i=0;i<table.size();i++)
if(table[i]!=i+1)
{
return i+1;
}
return table.size()+1;
}
int main()
{
int t;
cin>>t;
while(t--)
{
int n,m;
cin>>n>>m;
vector<int> s(m),t(m);
vector<set<int> > groups;
for(int i=0;i<m;i++)
cin>>s[i]>>t[i];
bool res = true;
for(int i=1;i<=n;i++)
{
vector<int> link;
int unlink = 0;
for(int j=0;j<m;j++)
{
if(t[j]==i)
link.push_back(s[j]);
if(s[j]>i)
break;
}
if(link.size()==i-1)//all n-1 is link,new set
{
set<int> a;
a.insert(i);
groups.push_back(a);
}
else if(link.size()<i-1)
{
unlink = findunlink(link,i);
set<int>* group;
for(int u=0;u<groups.size();u++)
{
if(groups[u].count(unlink)>0)
{
group=&groups[u];
break;
}
}
for(int j=1;j<i;j++)
{
if(find(link.begin(),link.end(),j)==link.end()&&(*group).count(j)==0)
{
cout<<"No"<<endl;
i=n+1;
res=false;
break;
}
else if(find(link.begin(),link.end(),j)!=link.end()&&(*group).count(j)>0)
{
cout<<"No"<<endl;
i=n+1;
res=false;
break;
}
}
if(res)
{
(*group).insert(i);
}
}
}
if(res)
cout<<"Yes"<<endl;
}
return 0;
}
c++第一题ac
#include <iostream>
#include <vector>
#include <set>
#include <algorithm>
using namespace std;
int findunlink(vector<int>& table,int n)
{
for(int i=0;i<table.size();i++)
if(table[i]!=i+1)
{
return i+1;
}
return table.size()+1;
}
int main()
{
int t;
cin>>t;
while(t--)
{
int n,m;
cin>>n>>m;
vector<int> s(m),t(m);
vector<set<int> > groups;
for(int i=0;i<m;i++)
cin>>s[i]>>t[i];
bool res = true;
for(int i=1;i<=n;i++)
{
vector<int> link;
int unlink = 0;
for(int j=0;j<m;j++)
{
if(t[j]==i)
link.push_back(s[j]);
if(s[j]>i)
break;
}
if(link.size()==i-1)//all n-1 is link,new set
{
set<int> a;
a.insert(i);
groups.push_back(a);
}
else if(link.size()<i-1)
{
unlink = findunlink(link,i);
set<int>* group;
for(int u=0;u<groups.size();u++)
{
if(groups[u].count(unlink)>0)
{
group=&groups[u];
break;
}
}
for(int j=1;j<i;j++)
{
if(find(link.begin(),link.end(),j)==link.end()&&(*group).count(j)==0)
{
cout<<"No"<<endl;
i=n+1;
res=false;
break;
}
else if(find(link.begin(),link.end(),j)!=link.end()&&(*group).count(j)>0)
{
cout<<"No"<<endl;
i=n+1;
res=false;
break;
}
}
if(res)
{
(*group).insert(i);
}
}
}
if(res)
cout<<"Yes"<<endl;
}
return 0;
}
java.util.*;
Main {
main(String[] args) {
Scanner sc = Scanner(System.);
(sc.hasNext()) {
n = sc.nextInt();
[] a = [n];
[] b = [n];
[] c = [n];
[] d = [n];
num = ;
(i=;i<n;i++){
a[i] = sc.nextInt();
b[i] = sc.nextInt();
c[i] = sc.nextInt();
}
(i=;i<n;i++){
(j=i+;j<n;j++){
(a[j]>a[i]){
(b[j]>b[i]){
(c[j]>c[i]){
d[i] = ;
}
}
}{
(a[j]<a[i]){
(b[j]<b[i]){
(c[j]<c[i]){
d[j] = ;
}
}
}
}
}
}
(i=;i<n;i++){
(d[i]==){
num++;
}
}
System..println(num);
}
}
}100%通过 顶部输入获取略过。就是个排序找中位数。对了，不能使用快排，因为这个傻瓜编译器认为快排比原生排序sort耗时更久(原生sort是冒泡排序，我就呵呵)。
var arr = ['3', '2', '9', '10', '11'];
var newArr = arr.map(function(i){
return parseInt(i);
});
var tempArr = newArr.sort(function(a,b) {
return a - b;
});
function findMid(arr) {
var len = arr.length;
return len % 2 === 0 ? (arr[len/2-1]+arr[len/2])/2 : arr[(len-1)/2];
}
var needNum = findMid(tempArr);
console.log(needNum);
第一题；通过率100%
function sortNum(a,b){
return a-b;
}
numSize = read.line();
numArr = read.line();
numArr = str.split(' ');
numArrNew = numArr.sort();
for(var i =0;i<numArr.length;i++){
numArrNew[i] = parseInt(numArr[i]);
}
numArrNew.sort(sortNum);
if(numArrNew.length%2 == 0){
zhongweishuS = parseInt(numArrNew.length/2);
zhongweishuX = zhongweishuS - 1;
zhongweushu = ( parseInt(numArrNew[zhongweishuS]) + parseInt(numArrNew[zhongweishuX]) )/2;
print(zhongweushu);
}else{
zhongweishu = Math.floor(parseFloat(numArrNew.length/2));
print(numArrNew[zhongweishu]);
}
第二题个人只做对了 27%，最后还有个判断那点没完善，正准备补充就提交了，只差一分钟，唉。
个人思路，找出重复次数最多的两项，不是同一项并且出现次数都大于1，都减1相乘得到最大利益。
话不多说，上代码，有个很简单的判断没写完，表示撸炸了不想写了，该去吃鸡压压惊，我写个注释，你们自己完善吧。
var n = 4;
var arr = [1, 2, 3, 4];
// 寻找重复数字和重复次数
function saveAgainCount(arr) {
var newArr = unique(arr);
var countArr = [];
newArr.forEach(function(item){
var tempArr = arr.filter(function(num){
return item === num;
});
var count = tempArr.length;
countArr.push({item, count});
});
return countArr;
}
// 数组去重，没办法，编译器不能使用ES6的set结构去重，只能这么去重了
function unique(arr) {
var res = [];
arr.forEach(function(e, i, arr) {
if (arr.indexOf(e) === i) {
res.push(e);
}
});
return res;
}
// 根据对象数组的重复次数排序
function compare(property) {
return function (obj1, obj2) {
var value1 = obj1[property];
var value2 = obj2[property];
return value1 - value2;
}
}
var finalArr = saveAgainCount(arr);
finalArr.sort(compare("count"));
/*请完成下面这个函数，实现题目要求的功能
当然，你也可以不按照下面这个模板来作答，完全按照自己的想法来 ^-^
******************************开始写代码******************************/
// 取最大重复次数俩值
function solve(arr) {
var len = arr.length;
var fir = arr[len-1];
var sec = arr[len-2];
/*
这里的判断还没完善，如果总城市不大于等于4，最大利益为0；如果出现的次数都不大于1，则最大利益为1。
*/
// 如果俩数组对象相同，证明没办法获取最大利润——修的路有交叉点
if (n < 4) {
return 0;
}
if (n) {
// 如果出现的次数都不大于1，则最大利益为1
// ...
}
// 最大重复次数-1再相乘
return (fir.count-1) * (sec.count-1);
}
/******************************结束写代码******************************/
var res;
res = solve(finalArr);
console.log(res);第一题：100% AC, 第二题放弃。
java.util.*Main {
Node{
ArrayList<Node> (value){
.= value= ArrayList<>()= }
}
(Scanner in){
n = in.nextInt()m = in.nextInt()HashMap<IntegerNode> map = HashMap<>()(i = i < ni++){
map.put(i+Node(i+))}
(i = i < mi++){
f = in.nextInt()s = in.nextInt()Node nf = map.get(f)Node ns = map.get(s)nf..add(ns)ns..add(nf)}
Node n1 = map.get()map.remove()List<Node> xl = n1.List<Node> nxl = ArrayList<>()(Map.Entry<IntegerNode> entry : map.entrySet()) {
(!xl.contains(entry.getValue())){
nxl.add(entry.getValue())}
}
pan = (Node node : nxl){
(Node node1: xl){
(!node..contains(node1)){
pan = }
}
}
(pan){
System..println()}{
System..println()}
}
(String[] args){
Scanner in = Scanner(System.)n = in.nextInt()(i = i < ni++){
(in)}
}
}java.util.*Main {
Node{
ArrayList<Node> (value){
.= value= ArrayList<>()= }
}
(Scanner in){
n = in.nextInt()m = in.nextInt()HashMap<IntegerNode> map = HashMap<>()(i = i < ni++){
map.put(i+Node(i+))}
(i = i < mi++){
f = in.nextInt()s = in.nextInt()Node nf = map.get(f)Node ns = map.get(s)nf..add(ns)ns..add(nf)}
Node n1 = map.get()map.remove()List<Node> xl = n1.List<Node> nxl = ArrayList<>()(Map.Entry<IntegerNode> entry : map.entrySet()) {
(!xl.contains(entry.getValue())){
nxl.add(entry.getValue())}
}
pan = (Node node : nxl){
(Node node1: xl){
(!node..contains(node1)){
pan = }
}
}
(pan){
System..println()}{
System..println()}
}
(String[] args){
Scanner in = Scanner(System.)n = in.nextInt()(i = i < ni++){
(in)}
}
}import java.util.*;
/*第一题 100% AC*/
public class Main {
public static class Node{
public int value;
public ArrayList<Node> nexts;
public boolean pass;
public Node(int value){
this.value = value;
nexts = new ArrayList<>();
pass = false;
}
}
public static void process(Scanner in){
int n = in.nextInt();
int m = in.nextInt();
HashMap<Integer, Node> map = new HashMap<>();
for(int i = 0; i < n; i++){
map.put(i+1, new Node(i+1));
}
for(int i = 0; i < m; i++){
int f = in.nextInt();
int s = in.nextInt();
Node nf = map.get(f);
Node ns = map.get(s);
nf.nexts.add(ns);
ns.nexts.add(nf);
}
Node n1 = map.get(1);
map.remove(1);
List<Node> xl = n1.nexts;
List<Node> nxl = new ArrayList<>();
for (Map.Entry<Integer, Node> entry : map.entrySet()) {
if(!xl.contains(entry.getValue())){
nxl.add(entry.getValue());
}
}
boolean pan = false;
for(Node node : nxl){
for(Node node1: xl){
if(!node.nexts.contains(node1)){
pan = true;
break;
}
}
}
if(pan){
System.out.println("No");
}else{
System.out.println("Yes");
}
}
public static void main(String[] args){
Scanner in = new Scanner(System.in);
int n = in.nextInt();
for(int i = 0; i < n; i++){
process(in);
}
}
}
第二题是先把ABC,分别放到不同的集合中,然后判断,如果是就直接remove,一次就过了....然鹅第一题不会....
public static void main(String[] args) {
// TODO Auto-generated method stub
int lowCount=0;
Scanner s1=new Scanner(System.in);
List<Integer> a1=new ArrayList();
List<Integer> a2=new ArrayList();
List<Integer> a3=new ArrayList();
int count=Integer.parseInt(s1.nextLine());
for (int i = 0; i < count; i++) {
String io2 = s1.nextLine();
String[] str2 = io2.split("\\s+");
a1.add(Integer.parseInt(str2[0]));
a2.add(Integer.parseInt(str2[1]));
a3.add(Integer.parseInt(str2[2]));
}
for (int i = a1.size()-1 ; i >=0;i--) {
for (int j = 0; j <a1.size(); j++) {
if(a1.get(i) < a1.get(j) && a2.get(i) < a2.get(j) && a3.get(i) < a3.get(j)){
lowCount++;
a1.remove(i);
a2.remove(i);
a3.remove(i);
break;
}
}
}
System.out.println(lowCount);
}
;
;
{
([] ) {
scanner= (.);
num = scanner.();
[][] things = [num][];
(i=;i<num;i++){
scanner1 = (.);
line = scanner1.();
[] lines = line.();
(j=;j<;j++){
things[i][j] =.(lines[j]);
}
}
res = (num,things);
..(res);
}
(,[][] ){
res = ;
count=;
bgm:(i = ; i <-; i++) {
(j = i+; j <; j++) {
count++;
([i][]<[j][]&&[i][]<[j][]&&[i][]<[j][]){
res++;
bgm;
}
}
}
..(+count);
res;
}
}// n 记录出入的数组总的个数
var n = readInt()
// data 存储输入的数组
var data =[]
for(var i=0; i<n; i++){
data.push(readInt())
}
// 选择排序 从小到大
for(var i=0; i<n; i++)
for(var j=i; j< n; j++){
if(data[i] > data[j]){
var temp = data[i]
data[i] = data[j]
data[j] = temp
}
}
var sum = 0;
for(var i=0; i<n; i++){
sum = sum + data[i];
}
if(n == 1){
sum = data[0]
} else{
if(n%2 != 0){
sum = data[Math.floor((n-1)/2)]
} else{
sum = parseFloat((data[Math.floor((n-1)/2)] + data[Math.floor((n-1)/2+1)])/2);
}
}
print(sum)
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