If you're stuck think like this, having a directed segment AB, you just need a C among the remaining points s.t. <CBA is acute. Fixing A, is there a way to choose B such that any C among remaining will form acute <CBA ?

That is test 2


Edited by author 14.11.2018 06:18

assc

Edited by author 20.03.2019 19:54

My code gives true answers for all tests but when I send problem it gives WA in test 3. please write answer.

#include <bits/stdc++.h>
using namespace std;
#define int long long
signed main() {
    ios_base::sync_with_stdio(false);
    cout.tie(0);
    cin.tie(0);
    int n,k;
    cin >> n >> k;
    int dontused=0,dontdie=0;
    int f;
    for(int i=0;i<n;i++)
    {
        cin >> f;
        if(k>f)
        {
            dontdie+=k-f;
        }
        if(k<f)
        {
            dontused+=f-k;
        }
    }
    cout << dontused << " " << dontdie;
    return 0;
}

What is the problem with my code?

#include <iostream>
using namespace std;
#include <cmath>

bool check_prime(double n) {
    int c;
    for (c=2;c<= sqrt(n);c++) {
        if((int) n%c==0)
            return 0;
    }
    if(c==(int) n)
        return 1;
}

int nth_prime(int n){
    int m=1;
    while (n>0) {
        m++;
        if (m == 2) {
            n--;
        }
        if (m%2!=0) {
            if (check_prime(m)==1) {
                n--;
            }
        }
    }
    return m;

}

int main()
{
    int n,m,i,a[15000],s;
    scanf("%d",&n);
    for(i=0;i<n;i++){
        scanf("%d",&a[i]);
    }
    for(i=0;i<n;i++){
        s = nth_prime(a[i]);
        printf("%d\n",s);
    }
    return 0;
}

a=input()
c=0
c+=a.count('a')
c+=a.count('d')
c+=a.count('g')
c+=a.count('j')
c+=a.count('m')
c+=a.count('p')
c+=a.count('s')
c+=a.count('v')
c+=a.count('y')
c+=a.count('y')
c+=a.count('.')
c+=a.count(' ')

c+=a.count('b')*2
c+=a.count('e')*2
c+=a.count('h')*2
c+=a.count('k')*2
c+=a.count('n')*2
c+=a.count('q')*2
c+=a.count('t')*2
c+=a.count('w')*2
c+=a.count('z')*2
c+=a.count(',')*2

c+=a.count('c')*3
c+=a.count('f')*3
c+=a.count('i')*3
c+=a.count('l')*3
c+=a.count('o')*3
c+=a.count('r')*3
c+=a.count('u')*3
c+=a.count('x')*3
c+=a.count('!')*3
print(c)

I can't understand problem in my algorithm. Can it be solved with just Dijkstra?

#include<stdio.h>
int main()
{
    int n, sum=0, i;

    scanf("%d", &n);
    if(n>0)
    {
        for(i=1; i<=n; i++)
            sum=sum+i;
    }
    else if(n<0)
    {
        for(i=n; i<=1; i++)
            sum=sum+i;
    }
    printf("%d", sum);

    return 0;
}

Isn't it?

помогите понять условие, пожалуйста, буду благодарен

#include <iostream>
#include <math.h>
using namespace std;
int main()
{ int n,k,i,a;float z;
cin>>n;
int *b;
b=new int [n];
for(i=1;i<=n;i++){cin>>k;z=sqrt((k-1)*8+1);a=z;if(a==z)b[i-1]=1;else b[i-1]=0;}
  for(i=1;i<=n;i++)cout <<b[i-1]<<" ";

    return 0;
}

import math
a=input().split()
for i in range(len(a)):
    u=int(a[i])
    l=str(math.sqrt(u))+'0000'
    for x in range(len(l)):
        if l[x]=='.':
            print(l[0:x+5])

Edited by author 10.11.2018 00:35

Buning uchun rand() funksiyasidan foydalaning.
masalan:
1000000 ta ababababbabab.....asdbasdsajkdbksajbdjkasbdkjbskfjbaksjdbfkjsad.
faqat ixtiyoriy ravishda bo'lsin

k = rand()%26 + 97 ;
qilib chiqaring shuni uzi kifoya

Edited by author 15.06.2013 11:32

intxt = input()

for i in intxt.split()[::-1]:
    print('%.4f' % int(i) ** 0.5)

Use the priority queue data structure.

input:
3 3
1 1 1
0
0 2
1 3 1 2
1 1 1 3
0
0 0
1
2 1
0 0

output:
-
-
1

#include <iostream>
#include <string>
using namespace std;
int main()
{
    string a[3000];
    int n;
    int m;
    cin >> n;
    cin >> m;
    for (int i = 0; i < n; i++)
    {
        cin >> a[i];
    }
    int j = m;
    for (int i = 0; i < 10; i++)
    {
        cout << a[j];
        j++;
        if (j == n )
        {
            j = 0;
        }
    }
    cout << endl;
    return 0;
}

I THINK THAT MY PROGRAM MUST HAVE AC BUT WA3
WHAT THE 3 TEST????
PLEASE POST TEST №3!!!!!!!