/*
Le Van Manh
Hanoi University Of Science And Technology
Viet Nam
*/
#include <iostream>
#include <stdio.h>
#include <string>
using namespace std;
int readAddressToValue(){
    int v = 0;
    string a;
    cin>>a;
    if(
        a=="Alice"||a=="Ariel"||a=="Aurora"||
        a=="Phil"||a=="Peter"||a=="Olaf"||
        a=="Phoebus"||a=="Ralph"||a=="Robin"){
        v = 0;
    }else
    if(
        a=="Bambi"||a=="Belle"||a=="Bolt"||
        a=="Mulan"||a=="Mowgli"||a=="Mickey"||
        a=="Silver"||a=="Simba"||a=="Stitch"){
        v = 1;
    }else{
        v = 2;
    }
    return v;
}
int main(){

    int address0;
    int address1;
    int steps = 0;
    int n;
    cin>>n;

    address0 = readAddressToValue();
    steps+= address0 - 0;
    for(int i=1;i<n;i++){
        address1 = readAddressToValue();
        steps += abs(address1 - address0);
        address0 = address1;
    }

    cout<<steps;
    return 0;
}

Edited by author 20.02.2016 13:17

You just need an array with 100 elements so in every position M, you add the ID using a dynamic structure such as "vector" in C++. (Theoretical complexity O(n)) Now you go from 100 to 0 and then print every element from each position M.

Edited by author 23.04.2016 15:06

if you have a better algo please tell to me.

plz help!!!!!!!!!!!!!!!

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main()
{
    int N,M,a,b;
    vector <int> t;
    vector <int> s;
    vector <int> st;
    cin>>N;
    for(int i=0; i<N; i++)
    {
        cin>>a;
        t.push_back(a);
    }
    cin>>M;
    for(int i=0; i<M; i++)
    {
        cin>>b;
        s.push_back(b);
    }
    for(int i=0; i<M; i++)
    {
        for(int j=0; j<N; j++)
        {
            if(s[i]==t[j])  st.push_back(s[i]);
        }
    }
    sort(st.begin(),st.end());
    cout<<st.size();
    return 0;
}

import java.util.Scanner;

public class SMSspam2 {

    public static void main(String[] args) {

        Scanner in = new Scanner(System.in);
        String s =in.nextLine();
        char a[] = s.toCharArray();

        int count = 0;
        for(int i=0;i<s.length();i++){
            count = count+getCount(a[i]);
        }
        System.out.println(count);
        in.close();
    }

    private static int  getCount(char a) {
        int count =0;
        if (a == 'a' || a == 'd' || a == 'g' || a== 'j' || a== 'm' || a == 'p' || a == 's' || a== 'v' || a == 'y'
                || a== '.' || a== ' ') {
            count++;
        }
         if (a== 'b' || a == 'e' || a== 'h' || a == 'k' || a == 'n' || a == 'q' || a== 't' || a== 'w' || a== 'z'
                || a == ',') {
            count = count + 2;
        }
        if (a== 'c' || a== 'f' || a == 'i' || a == 'l' || a== 'o' || a== 'r' || a == 'u' || a== 'x' || a == '!') {
            count = count+ 3;
        }
        return count;
    }
}

#include <stdio.h>


int main()
{
#ifndef ONLINE_JUDGE
    freopen("input.txt", "rt", stdin);
    freopen("output.txt", "wt", stdout);
#endif

    int n;
    scanf("%d", &n);

    long *sections = new long[n];
    for (int i = 0; i < n; i++){
        scanf("%ld", &sections[i]);
    }

    if (n==3){
        printf("%ld %d", sections[0] + sections[1] + sections[2], 2);
    }
    else{
        int i = 0;
        int maxpoint = i;
        while (i + 3 < n){
            int j = i + 1;
            if (sections[i] + sections[i+1] + sections[i+2] < sections[j] + sections[j+1] + sections[j+2]){
                maxpoint = j;
            }
            i++;
        }
        printf("%ld %d", sections[maxpoint] + sections[maxpoint+1] + sections[maxpoint+2], maxpoint+2);
    }

    return 0;
}

var a, x: extended;
begin
readln (a);
x:= sqrt (a);
writeln (x);
end.

Выдаёт неправильный ответ.

import java.util.Scanner;

public class Digitals {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int N = in.nextInt();
        int a[] = new int[N];
        int b[] = new int[N];
        int sum = 0;
        for (int i = 0; i < N; i++) {
            a[i] = in.nextInt();
            sum = 1;
            for (int j = 0;; j++) {
                sum = sum + j;
                if (sum == a[i]) {
                    b[i] = 1;
                    break;
                }
                if (sum > a[i]) {
                    b[i] = 0;
                    break;
                }
            }
        }
        for (int p = 0; p < N - 1; p++)
            System.out.print(b[p] + " ");
        System.out.println(b[N - 1]);
        in.close();
    }
}

How do it more fast?

#include <iostream>
using namespace std;

int main()
{
    int x, y, c;

    cin >> x >> y >> c;

    if((x + y) < c)
    {
        cout << "Impossible" << endl;
    }
    else
    {
        for(int i = 0; i <= x; i++)
        {
            for(int j = 0; j <= y; j++)
            {
                if((i + j) == c)
                {
                    cout << i << " " << j << endl;
                    return 0;
                }
            }
        }
    }
}

Edited by author 13.10.2015 20:23

Edited by author 13.10.2015 20:23

There are 21 steps in this sequence:

1, 2, 3, 4, 5, 6, out, 6, 7, 8, 9, 10, 11, 12, out, 12, 11, 12, 13, 14, 15

a = int(input())
b = int(input())
print (a+b)
_____________________
почему не правильно пишет?

I'm using BS + Dijkstra but I'm getting TLE, why? I optimized solution by using edges that have capacity >= 3 * 10^6, implemented BS with iterations (without "while"), but keep getting TLE.

what if k=7?
Is l=2?

Just list all of cases like x>=y>=c or y>=x>=c
and you'll understand

//1152
#include <iostream>
#include <vector>
#include <cmath>


using namespace std;

int main() {
    int n, x0, y0, index0;
    cin >> n;
    if (n == 2){
        cout << 1 + ' ' + 2;
    } else {
        vector<vector<int>> p(n, vector<int>(2)); // [x, y]
        vector<double> k(n);
        for (int i = 0; i < n; i++) {
            cin >> p[i][0] >> p[i][1];
        }

        x0 = y0 = 2147483647;
        for (int i = 0; i < n; i++) {
            if (p[i][0] < x0 ||p[i][0] == x0 && p[i][1] < y0){
                x0 = p[i][0];
                y0 = p[i][1];
                index0 = i;
                }
        }

        for (int i = 0; i < n; i++) { //y = kx + m
            if (i == index0) {
                k[i] = 9223372036854775807;
            } else {
                if (x0 == p[i][0]) {
                    k[i] = 9223372036854775807;
                } else {
                    k[i] = (p[i][1] - y0) / (p[i][0] - x0);
                }
            }
        }


        vector<int> d(n);
        for (int i = 0; i < n; i++) {
            d[i] = i;
        }

        double tempD;
        int tempI;
        for (int i = n - 1; i >= 0; i--) {
            for (int j = 0; j < i; j++) {
                if (k[j + 1] < k[j]) {

                    tempD = k[j + 1];
                    k[j + 1] = k[j];
                    k[j] = tempD;

                    tempI = d[j + 1];
                    d[j + 1] = d[j];
                    d[j] = tempI;

                }
            }
        }
        cout << index0 + 1 << ' ' << d[n / 2  -1] + 1;
    }
return 0;
}

I wrote program on Python 3, but second test is wrong for me. Why? I need a hint. Maybe, you can wrote test for my program?

Idea:
1. sort length descending
2. take first that smaller SUM in row
3. reduce SUM and colored ship
4. take next
5. finally for SUM if they == 0 -> go to next row


My solution:

[n, m] = [ int(x) for x in input().split() ]
a = []
for i in range(n):
    a.append(int(input()))
b = []
for i in range(m):
    b.append(int(input()))
a.sort(reverse=True)
color = [-1 for x in range(n)]

i = 0
while i < m:
    s = b[i]
    z = 0
    while z < n:
        aa = []
        j = z
        while s > 0:
            while j < n and a[j] > s :
                j += 1
            if j >= n:
                for ee in aa:
                    s += a[ee]
                    color[ee] = -1
                aa = []
                break
            elif color[j] != -1:
                j += 1
            else:
                s = s - a[j]
                aa.append(j)
                color[j] = i
        z += 1
        if s == 0:
            break
    i += 1
for i in range(m):
    aa = []
    cnt = 0
    for j in range(n):
        if color[j] == i:
            aa.append(a[j])
            cnt += 1
    print(cnt)
    aa.sort(reverse=True)
    ab = [str(x) for x in aa]
    ss = " ".join(ab)
    print(ss)