why is  k =10 and n =2 output =90

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int main() {
unsigned long long k[4];
for(int i=0;i<4;i++){
    scanf("%llu",&k[i]);

}
printf("\n");
printf("%.4f\n",sqrt(k[3]));
printf("%.4f\n",sqrt(k[2]));
printf("%.4f\n",sqrt(k[1]));
printf("%.4f\n",sqrt(k[0]));
  return 0;
}

Ошибка на тесте 2 /Error in test 2

Edited by author 09.09.2018 15:45

Почините алгоритм проверки! Не контачит на сях. Вот код:
#include <stdio.h>

int main(int argc, char const *argv[])
{
    int a, b;
    a=1, b=5;
    printf("%i\n", a+b);
    return 0;
}

Hello, everyone.
Just wanted to summarize some info about this problem.
Ok, my code got AC with following things in it:
1. My pi was 3.1415926535897932384626433 (perhaps it is enough)
2. I used this formula from wikipedia
deltaArc=acos(sin(phiA)*sin(phiB)+cos(phiA)*cos(phiB)*cos(deltaL));
distance = deltaArc*3437.5;
(see the first formula from wikipedia http://en.wikipedia.org/wiki/Great-circle_distance)
3. I used the following condition:
if(100.00-distance>0.005) printf("DANGER!\n");
Hope it will help somebody.

1)Solution always exists and his complexity <= O(n^2)
2)Try to solve the problem when the permutation is n (n-1) ... (n-left+1) 1 2 3 ... right where
left+right=n.
   After O(n^2) steps you can get n 1 2 3 ... (n-1).
   Then after O(n^2) steps you can get 1 2 3 ... n.
3)Try to reduce the problem to this permutation by O(n^2) steps.
4)Good to know: 1 2 3 ... t x-> 2 1 3 ... t x -> 3 1 2 4 ... t x -> 4 1 2 3 5 ... t x ->... ->
t 1 2 3 ... (t-1) x -> x 1 2 3 ... t by t steps. I call this operation "Shift(t)".

Edited by author 26.07.2018 20:20

Edited by author 22.09.2018 14:02

I can't think of a way to store goto and fail transitions of so many vertices and avoid MLE. please help.

here's code:

#include <stdio.h>
#include <queue>
using namespace std;
short g[110000][256];
short f[110000];
short o[110000];
short n,m,newstate=0;
queue<short>Q;
int main(void){
    short i,pos=0,q,r,u,v,ch,state;
    scanf("%hd",&n);getchar();
    for(i=1;i<=n;++i){
        r=0;
        ch=getchar();
        state=0;
        ++r;
        ch+=128;
        while(g[state][ch]){
            state=g[state][ch];
            ch=getchar();
            ++r;
            ch+=128;
        }
        while(ch!='\n'+128){
            newstate=newstate+1;
            g[state][ch]=newstate;
            state=newstate;
            ch=getchar();
            ++r;
            ch+=128;
        }
        o[state]=r-1;
    }
    for(i=0;i<256;++i){
        if(q=g[0][i]){
            f[q]=0;
            Q.push(q);
        }
    }
    while(!Q.empty()){
        r=Q.front();
        Q.pop();
        for(i=0;i<256;++i){
            u=g[r][i];
            if(!(u==0&&r>0)){
                Q.push(u);
                v=f[r];
                while(g[v][i]==0&&v>0) v=f[v];
                f[u]=g[v][i];
                if(o[u]==0) o[u]=o[f[u]];
            }
        }
    }
    scanf("%hd",&m);getchar();
    for(i=1;i<=m;++i){
        q=0;r=0;
        while((ch=getchar())!='\n'){
            ch+=128;
            ++r;
            while(g[q][ch]==0&&q>0) q=f[q];
            q=g[q][ch];
            if(o[q]){
                printf("%hd %hd",i,r-o[q]+1);
                return 0;
            }
        }
    }
    printf("Passed\n");
    return 0;
}

For TC:
abracadabra
abrabracada

why don't the prefixes are:
abra abracada a?

and also

why for TC:
abracadabra
arbadacarba

the prefix is not:
a?

#include <bits/stdc++.h>
using namespace std;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n,count = 1;
    long long m, sum = 0;
    vector<long long> coll;
        vector<pair<long long, int>> coll1;
    cin >> n;
    while(n--)
    {
        cin >> m;
        coll.push_back(m);
    }
    m = coll[0];
    int i = 1;
    if(coll.size() > 3){
    for( ; i < coll.size(); ++i)
        {
           if(m <= coll[i] && count < 3)
       {
           count++;
       } else
           if( i>=3)
           {
               sum += coll[i-1];
               sum += coll[i-2];
               sum += coll[i-3];
               coll1.push_back({sum, i-1});
               sum = 0;
               count = 1;
           }
       m = coll[i];
       }
       sort(coll1.begin(), coll1.end());
       cout << coll1[coll1.size()-1].first << ' ' << coll1[coll1.size()-1].second  << endl;
    } else
    cout << accumulate(coll.begin(), coll.end(), 0) << ' '<<'2' << endl;
       coll.clear();
       coll1.clear();
       return 0;
}

It would be awesome to have stats for each task, like best/average time/memory of AC-ed solutions for each language.

4 0
0 0 0 0

Edited by author 23.05.2007 21:03

use  ios_base::sync_with_stdio(false); cin.tie(0); with cin/cout. Don't forget to add last two digit's sum carry if it has.

if you read data as:
double x;
cin>>x;
a[i]=x*100;
it doesn't work!

n = int(input())

l = list(map(int, input().split()))
l1 = []

for i in range(n):
    l1.append((l[i]//2 +1))

if n > 3:
    k = (n // 2 + 1) // 2 + 1 #Groups which decline
else:
    k = (n // 2 + 1) // 2

for i in range(k):
    l1.remove(max(l1))

print(sum(l1))


Edited by author 15.09.2018 21:04

x[i] - y[i] = a * i + b

x[i-1] - y[i-1] = a * (i-1) + b

------------------------------------

x[i] - x[i-1]  - ( y [i] - y[i-1] ) = a



Let x1[i] = x[i] - x[i-1], i > 0. and y1[i] = y[i] - y[i-1], for i>0.


so ,

x1[i] - y1[i] = a,  a = x1[0] - y1[0].

x1[i-1] - y1[i-1] = a
-------------------------------
x1[i]  - x1[i-1] - (y1[i ] - y1[i-1]) = 0  or,

x1[i] - x1[i-1] = y1[i] - y1[i-1]

Let x2[i] = x1[i] - x1[i-1] = x[i] - 2*x[i-1] + x[i-2], for i>1. and
    y2[i] = y1[i] - y1[i-1] = y[i] - 2*y[i-1] + y[i-2], for i>1.

so  x2[i] = y2[i],  for all 1 < i <= n.

Use KMP.

I didn't try this, yet ).

this may be true: 1.000 > 1.
so add this func:
    double acos1(double x)
    {
        if(x > 1)
          x = 1;
        if(x < -1)
          x = -1;
        return acos(x);
    }

1 4 1
1
1 2 3 4
Ans 5

6 5 50
1 4 6 6 2 5
2 5 10 14 15
Ans 21

6 5 4
1 4 6 6 2 5
2 5 10 14 15
Ans 13

9 5 4
2 4 2 5 3 1 7 2 3
5 6 9 10 14
Ans 22

just use dijkstra and that's all