Hi, folks!

Can anyone give me the source code in C# of this task (mine is already AC) when the timing is less than 0.1 sec?
Or can anyone give me the key notes how it can be decreased in almost ten times?

Thanks in advance!

#include <iostream>
using namespace std;

int main() {
    int cand, ch;
    cin >> cand >> ch;
    int* Ch = new int[ch];
    double* Cand = new double[cand];

    for (int i = 0; i < cand; i++) {
        Cand[i] = 0;
    }

    for (int i = 0; i < ch; i++)
        cin >> Ch[i];

    for (int i = 0; i < ch; i++) {
            Cand[Ch[i]-1] += 1;
    }

    for (int i = 0; i < cand; i++) {
        printf("%.2f%%\n", (Cand[i]/6.0)*100.0);
    }
}
+++++++++++
Why this isn't work?

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

int main() {
    int i,j,N,p;
    std::cin>>N;

    float cd[N][2],R,dist = 0.0;
    std::cin>>R;

    for(p = 0;p < N;p++)
        std::cin>>cd[p][0]>>cd[p][1];

    i = 0;
    j = i + 1;

        dist = (2 * std::acos(-1.0) * R);

    while(i < N)
    {
        dist += std::sqrt( ( (cd[j][0] - cd[i][0]) * (cd[j][0] - cd[i][0]) ) +
( (cd[j][1] - cd[i][1]) * (cd[j][1] - cd[i][1]) ) );
        i++;
        j = (i + 1) % N;
    }

    std::cout <<std::setprecision(2) << std::fixed <<dist;
    return 0;
}

What answer should be for given test?

1 1 1 1
0

Can anyone explain the sample:

3
0 0 0
1 1 0
1 1 0



1
1 3


I dont understand it. The resulting matrix is:
0 1 1
0 1 1
0 0 0

Thanks!

Same solution:

Visual C++ 2013:
Accepted    0.686    8 120 КБ

G++ 4.9 C++11:
Time limit exceeded    21    1.513    3 924 КБ

oh my code..

first  use dynamic programming to find the minimum value of distance from 0 and visit 1~m once then to 0, we can get a chain 0--i1-i2--...-im--0

then we consider to cut this chain, use dynameic programming dp[i] to record distance.we cut first i node of the chain..
dp[m] is the result..

I'll try this idea...

rt

can someone explain test number 2?

please post it (read chunk) here or give some piece of code, many sanx!

#include <stdio.h>
#include <math.h>

int main( int argc, char ** argv )
{
    unsigned long number[256*1024];
    int counter = 0;
    while( scanf("%lu",&number[counter]) != EOF )
    {
        counter++;
    }

    for( counter--; counter >= 0; counter-- )
        printf("%.4f\n",sqrt(number[counter]));
    return 0;
}

i just found no of storngly connected components and got ac in 0.562 sec after eliminating stack overflow with pragma.

how to make it so fast as 0.031 sec and so little memory?

Try this test:

Input:
5 1
100 0
100 0
100 0
100 0
100 0
2 50

Output:
250

Hello! I have strange WA 22 and really don't understand why. Can anyone help to find the bug in my code. (corrected)

Edited by author 04.12.2016 20:25

import java.util.*;
import java.io.*;

public class MEGA
{
    public static void main(String arg[])
    {
        int k,n,i,sum=0;
        Scanner scan=new Scanner(System.in);
        k=scan.nextInt();
        n=scan.nextInt();
        int[] a=new int[n];
        for(i=0;i<n;)
        {
            if(scan.hasNextInt())
            {
                a[i]=scan.nextInt();
                i++;
            }
        }
        for(i=0;i<n;i++)
        {
            sum+=a[i]-k;
        }
        if(sum<0)
        {
            System.out.println("1");
        }
        else
        {
            System.out.println(""+sum);
        }
    }
}

This very simple program is turning out to be a headache. What is Test #3?

ll n;
ll ar[300][300];
int main(){
    cin >> n;
    for(ll i = 1; i <= n; i ++) for(ll j = 1; j <= n; j ++) ar[i][j] = 0;
    for(ll i = 1; i <= n; i ++){
        while(1){
            ll x; cin >> x;
            if(x == -1) break;
            ar[x][i] = 1;
        }
    }
    for(ll i = 1; i <= n; i ++) ar[i][n+1] = 1;
    ll col = 1;
    while(col <= n){
        if(ar[col][col] == 1){
            for(ll i = 1; i <= n; i ++){
                if(i == col) continue;
                if(ar[i][col] == 1){
                    //choose ith row
                    for(ll j = 1; j <= n+1; j ++) ar[i][j] = ar[i][j]^ar[col][j];
                }
            }
        }
        else{
            break;
        }
        col ++;
    }
    bool f = 1;
    ll br[300][300];
    for(ll i = 1; i <= n; i ++) for(ll j = 1; j <= n; j ++) if(i==j) br[i][j] = 1;
    for(ll i = 1; i <= n; i ++){
        for(ll j = 1; j <= n; j ++) if(ar[i][j]!=br[i][j]) f = 0;
    }
    if(!f) cout << "No solution" << endl;
    else{
        for(ll i = 1; i <= n; i ++){
            if(ar[i][n+1]==1) cout << i << " ";
        }
    }
    return 0;
}

I have used basic Gauss algorithm making the augmented matrix equal to an identity matrix. Still getting WA. Please help
Thank you :)

Edited by author 04.12.2016 10:11

"If the King finds that the Architect has used more resources to build the wall than it was absolutely necessary to satisfy those requirements, then the Architect will LOOSE his head."
http://grammarist.com/usage/loose-lose/

I got AC with a wrong solution during the contest, so I think the test cases might be a bit weak. One example my wrong solution didn't pass is:
7 63713822
It printed 3, while the best answer is 2: 7 -> 8 -> 31856912 -> 31856911 -> 63713822.

Can you please add it to the test-set?

(Edit: I've fixed my solution now, but others may fail.)

Edited by author 03.12.2016 18:42

my code
#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n, c = 0, s = 0, v = 0, maxn = 0;
    cin >> n;
    string a, b;
    for(int i = 1; i <= n; i++){
        cin >> a ;
        for(int i = 0; i <= a.size(); i++){
        if(a[0] == 'E' && a[1] == 'm' && a[2] == 'p' && a[3] == 'e' ){
            c++;
            break;
        }
        else if( a[0] == 'M' && a[1] == 'a' && a[2] == 'c' && a[3] == 'a'){
            s++;
            break;
        }
        else if(a[0] == 'L' && a[1] == 'i ' && a[2] == 't' && a[3] == 't' ){
            v++;
            break;
        }
        }
        cin >> b;
    }
    if(c > s && c > v){
        cout << "Emperor Penguin";
    }
    else if(s > c && s > v){
        cout << "Macaroni Penguin";
    }
    else if(v > c && v > s){
        cout << "Little Penguin";
    }
    return 0;
}
why incorrect

WA7
3
000
->
000000000002

WA10
0
->
000000000002