Отметь точки на по теме «система коордипеть точки на координатной плоскости. соедините точки по порядку.no1(0, 9), (3, 13), (7, 10), (10, 12), (13, 11), (12, 13), (10, 15), (12, 19), (15, 19),(17, 17), (17. 13) (1916), (23, 16), (24, 20), (27, 20), (25, 19), (24, 14), (21, 14). | (e(18; 10), (21, 8), (20, 4), (17, 4), (15, 3), (16, 5), (18, 5), (18, 7), (16, 9), (13, 9),(10, 10), (6, 9), (3, 11), (0, 9).(14, 14), (15, 15).(12, 16).(14, 17).no3(0,6), (2,8), (3,11), (6,13), (10,14), (14,17), (16,16), (2016), (22,17), (24,15),| (24,11), (21,9), (20,11), (20,7), (19,6), (16,6), (16,7), (18,7), (19,8), (17,10),(16, 10), (15,11), (14,9), (11,11), (10, 14).(18,6), (18,4), (16,0), (14,2), (15,4), (15,5), (14,5), (14,0), (11,0), (11,6), (6,6),(6,,1), (3,1), (4,5), (3,8), (0,6).
//Первый ввод - число, второй - система счисления
//Pascal ABC.NET v3.0
var
a,i,b,r,n,j,bug:integer;
s,se,slo,slof:string;
procedure preob(var a,b,n:integer; var se:string);
begin
repeat
b:=a mod n;
a:=a div n;
str(b,se);
s+=se;
until (a<=n-1);
end;
begin
readln(slo);
readln(n);
for j:=1 to length(slo) do
begin;
a:=ord(slo[j]);
preob(a,b,n,se);
str(a,se);
s+=se;
for i:=1 to length(s) div 2 do
begin;
se:=s[i];
s[i]:=s[length(s)-i+1];
s[length(s)-i+1]:=se[1];
end;
write(s,'-');
slof:=slof+s;
delete(s,1,length(s));
end;
end.
//Слово Программа она кодирует как 11001111-11110000-11101110-11100011-11110000-11100000-11101100-11101100-11100000-
#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <algorithm>
#include <functional>
#include <cstring>
#include <utility>
#include <bitset>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstdio>
#include <climits>typedef unsigned long long ulol;
typedef long double ld;
typedef long long lol;
typedef long int li;#define mp make_pair
#define F first
#define S second
#define sqr(a) ( (a) * (a) )
#define pb push_back
#define INF 999999999
#define ret(a) cout << endl; system("pause"); return(a)
//#define ret(a) return(a)using namespace std;int main()
{
ld x;
cin >> x;
x = ( 8 / sqrt( x ) ) + sqrt(x);
cout << x;
ret(0);
}