thanks

#include <bits/stdc++.h>

using namespace std;
float p,q,t1,t2,t3,l1,l2;
int main()
{
    cin>>p>>q;
    for(int i=1; i<=1000000; i++){
        t1=(p*(float)i)/(float)100;
        t2=(q*(float)i)/(float)100;
        l1=(int)t1;
        l2=(int)t2;
        if(l1==t1||l2==t2||l1!=l2){
            if(l1!=t1) l1=l1+1;
            for(int j=l1; j<=l2; j++){
                t3=((float)j/(float)i)*(float)100;
                if(p-0.000001<t3&&t3<q+0.000001){
                    cout<<i;
                    return 0;
                }
            }
        }
    }
}

I can only get

2 3 1
3 4 3
2 3 2
according to the sample output...

has anyone done it in O(n)?

if(a/2. >= r)
{
    printf("%.3lf", pi*r*r);
    return 0;
}
This is 4th test, use only "%.3lf" !

Don't forget about cargo delivery time:
4
1 2 1
2 3 10
2 4 1
------
6

Edited by author 11.02.2017 17:10

I hope somebody will explain me what's wrong with my program, cause I really dont understand why i get out of memory (900KB) when my program should use only about 700KB (no more than 750KB definetly!).
So pls help me smb. Now i try to explain my problem.
The only memory i use, i define out of main scope and i dont use any dynamic allocations. So, the definition looks like:

unsigned int length[MAXIMUMSTACKS];
unsigned int maxlength[MAXIMUMSTACKS];
unsigned int base[MAXIMUMSTACKS];
unsigned int stackPtr[MAXIMUMSTACKS];
unsigned char opbuf[OPSIZE * MAXOP];
unsigned int stack[MAXOP / 2];

where MAXIMUMSTACKS = 1000, OPSIZE = 5, MAXOP = 100000.

So, let's count all memory we may use: 4*1000 + 4*1000 + 4*1000 + 4*1000 + 500000 + 4*50000 = 16*1000 + 500000 + 200000 = 716000 bytes! I is about 700KB to be honest.
So, i dont inderstand, why my program get Memory Limit Error and use (as Timus says) 900KB(!). Pls, help me to fix it. Tell me what i dont understand. I give you the code of programm below.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MASK(k) ((1 << k) - 1)
#define OPSIZE 5
#define POPCODE 1000000001
#define MAXIMUMSTACKS 1000
#define MAXOP 100000

unsigned int N;
unsigned int length[MAXIMUMSTACKS];
unsigned int maxlength[MAXIMUMSTACKS];
unsigned int base[MAXIMUMSTACKS];
unsigned int stackPtr[MAXIMUMSTACKS];
unsigned char opbuf[OPSIZE * MAXOP];
unsigned int stack[MAXOP / 2];
char op[5];

unsigned short stnum;
unsigned int value;
unsigned int indx;


int main(int argc, char *argv[]) {

//    freopen("input.txt", "r", stdin);

    // init
    scanf("%u", &N);
    memset(opbuf, 0, N * OPSIZE);
    memset(length, 0, MAXIMUMSTACKS * sizeof(unsigned short));
    memset(maxlength, 0, MAXIMUMSTACKS * sizeof(unsigned short));

    // fill opbuf
    for (unsigned int i = 0; i < N; ++i) {

        scanf("%s%hu", op, &stnum);
        --stnum;

        indx = OPSIZE * i;

        opbuf[indx] = *((unsigned char*)&stnum);
        opbuf[indx + 1] = *((unsigned char*)&stnum + 1) & MASK(2);

        if (op[1] == 'U') {
            scanf("%u", &value);

            value = (value << 2);
            for (unsigned int j = 0; j < 4; ++j)
                opbuf[indx + 1 + j] += *((unsigned char*)&value + j);

            ++length[stnum];
            if (length[stnum] > maxlength[stnum]) {
                maxlength[stnum] = length[stnum];

                if (maxlength[stnum] > MAXOP / 2)
                    maxlength[stnum] = MAXOP / 2;
            }

        } else {
            value = (POPCODE << 2);
            for (unsigned int j = 0; j < 4; ++j)
                opbuf[indx + 1 + j] += *((unsigned char*)&value + j);

            --length[stnum];
        }
    }

    // create big memory block for stacks
    unsigned int prevLength = 0;
    unsigned int prevBase = 0;
    for (unsigned int i = 0; i < MAXIMUMSTACKS; ++i) {
        if (maxlength[i] == 0)
            continue;

        base[i] = prevBase + prevLength;
        stackPtr[i] = base[i];
        prevBase = base[i];
        prevLength = maxlength[i];
    }

    // process through opbuf
    for (unsigned int i = 0; i < N; ++i) {
        stnum = 0;
        value = 0;
        indx = OPSIZE * i;

        stnum += opbuf[indx];
        stnum += ((opbuf[indx + 1] & MASK(2)) << 8);

        for (unsigned int j = 0; j < 4; ++j)
            *((unsigned char*)&value + j) = opbuf[indx + 1 + j];

        value = ((value >> 2) & 0x3fffffff);

        if (value == POPCODE) {

            if (stackPtr[stnum] == base[stnum]) {
                stackPtr[stnum] = base[stnum] + maxlength[stnum] - 1;
            } else {
                stackPtr[stnum]--;
            }

            printf("%u\n", stack[stackPtr[stnum]]);
        } else {
            stack[stackPtr[stnum]] = value;

            if (stackPtr[stnum] == base[stnum] + maxlength[stnum] - 1)
                stackPtr[stnum] = base[stnum];
            else
                stackPtr[stnum]++;
        }
    }

    return 0;
}

Edited by author 20.08.2015 16:44

Why i'm getting WA#1?

#include<bits/stdc++.h>
using namespace std;

int capacities[2002][2002];
int flowPassed[2002][2002];
vector<int> graph[2002];
int parentsList[2002];
int currentPathCapacity[2002];

int bfs(int startNode, int endNode)
{
    memset(parentsList, -1, sizeof(parentsList));
    memset(currentPathCapacity, 0, sizeof(currentPathCapacity));

    queue<int> q;
    q.push(startNode);

    parentsList[startNode] = -2;
    currentPathCapacity[startNode] = 999;

    while(!q.empty())
    {
        int currentNode = q.front();
        q.pop();

        for(int i=0; i<graph[currentNode].size(); i++)
        {
            int to = graph[currentNode][i];
            if(parentsList[to] == -1)
            {
                if(capacities[currentNode][to] - flowPassed[currentNode][to] > 0)
                {
                    parentsList[to] = currentNode;
                    currentPathCapacity[to] = min(currentPathCapacity[currentNode],
                    capacities[currentNode][to] - flowPassed[currentNode][to]);
                    if(to == endNode)
                    {
                        return currentPathCapacity[endNode];
                    }
                    q.push(to);
                }
            }
        }
    }
    return 0;
}

int edmondsKarp(int startNode, int endNode)
{
    int maxFlow = 0;
      while(true)
    {
        int flow = bfs(startNode, endNode);
        if (flow == 0)
        {
            break;
        }
        maxFlow += flow;
        int currentNode = endNode;
        while(currentNode != startNode)
        {
            int previousNode = parentsList[currentNode];
            flowPassed[previousNode][currentNode] += flow;
            flowPassed[currentNode][previousNode] -= flow;
            currentNode = previousNode;
        }
    }
    return maxFlow;
}

int main()
{
    long int m,n,k,x;

    cin>>m>>n>>k;
    x=m+n;
    for(int i=0;i<k;i++)
    {
        int a,b;
        cin>>a>>b;
        graph[a].push_back(m+b);
        graph[m+b].push_back(a);
        capacities[a][m+b]=1;
    }
    for(int i=0;i<m;i++) {graph[0].push_back(i+1);capacities[0][i+1]=1;}
    for(int i=0;i<n;i++) {graph[m+i+1].push_back(x+1);capacities[m+i+1][x+1]=1;}
    int y=edmondsKarp(0,x+1);
    int p=(m-y+1)/2;
    p=p+(n-y+1)/2;
    cout<<(y+p)<<endl;

}

import java.text.DecimalFormat;
import java.util.Scanner;

/**
 * Created by user on 06.02.2017.
 */
public class nytka {
    public static void main(String[] args) {
        double c = 0;
        Scanner sc = new Scanner(System.in);
        String a = sc.next();
        int b = Integer.valueOf(a);
        String q = sc.next();
        double r = Double.valueOf(q);
        double[] x = new double[b];
        double[] y = new double[b];
        if (100>b) {
            if (b > 0){
                for (int i = 0; i <= b - 1; i++) {
                    String x2 = sc.next();
                    double x1 = Double.valueOf(x2);
                    x[i] = x1;
                    String y2 = sc.next();
                    double y1 = Double.valueOf(y2);
                    y[i] = y1;

                }
                for (int i = b - 1; i > 0; i--) {
                    double rez = Math.sqrt(((x[i] - x[i - 1]) * (x[i] - x[i - 1])) + ((y[i] - y[i - 1]) * (y[i] - y[i - 1])));
                    c = c + rez;
                }
            }
        }
        double end = Math.sqrt(((x[b-1]-x[0])*(x[b-1]-x[0]))+((y[b-1]-y[0])*(y[b-1]-y[0])));
        c = c + end;
        double rad = (Math.PI*r*r)*(b/2);
        double endrez = c + rad;
        DecimalFormat f = new DecimalFormat("#,##0.00");
        System.out.print(f.format(endrez));
    }
}

using System;
using System.Threading;
using System.Globalization;

namespace TestOfCSharp
{
    class Program
    {
        static void Main(string[] args)
        {
            Thread.CurrentThread.CurrentCulture = CultureInfo.InvariantCulture;
            string[] input = Console.ReadLine().Split(' ');
            int N = int.Parse(input[0]);
            int R = int.Parse(input[1]);
            double sum = 0;
            for (int i = 0; i < N; i++)
            {
                input = Console.ReadLine().Split(' ');
                sum += double.Parse(input[0]) + double.Parse(input[1]);
            }
            sum += 2 * Math.PI * R;
            Console.WriteLine("{0:F2}", sum);
        }
    }
}

Edited by author 20.12.2013 05:18

Can somebody give me test #4? I'm getting WA

Don't use cin(TL1) and getc(WA1), use scanf and get AC:

char s[(int)1e7];
scanf("%s", &s);
length = strlen(s);

in 1 test case we can't determine whether subject 3 should be earlier than 4 or 4 than 3, because there's isn't any edge that show which one should be sdutied ealrier, why answer is YES???? I think if we can't verify it answer should be NO

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

int calculate (queue<int> wind)
{
        int result = 0;
        for (int i = 0; i < 3; i++) {
                result += wind.front();
                wind.pop();
        }
        return result;
}

int main() {
        int n, sum = 0, max=0, el, position=2, middle=2;
        queue<int> window;

        cin >> n;
        for (int i = 0; i < 3; i++) {
                cin >> el;
                window.push(el);
                max += el;
        }
        for (int i = 0; i < n-3; i++) {
                cin >> el;
                window.push(el);
                window.pop();
                sum = calculate(window);
                position++;
                if(sum > max) {
                        max = sum;
                        middle = position;
                }
        }
        cout << max << " " << middle;
        return 0;
}

import java.util.*;
public class JavaApp {

    public static void main(String[] args) {
             int i=0;

                Scanner scan = new Scanner(System.in);
        int n=scan.nextInt();

                if ((n>=1)&&(n<10000)){
                    i=n *(n + 1) / 2;
                    System.out.print(i);}
                else
                if ((n<=1)&&(n>-10000)){
                    i=-((-n) *(1 -n) / 2) + 1;
                    System.out.print(i);}
        };
    }

This is the right solution
Sorry for my english

var ans: array[1..1000] of longint;
n,m,y,q,i: longint;

function BinPow(x,y: longint): longint;
begin
  if y=1 then BinPow:=x
  else if y mod 2=0 then BinPow:=sqr(BinPow(x,y div 2)) mod M
  else BinPow:=(sqr(BinPow(x,y div 2))*x) mod M;
end;

begin
  readln(n,m,y);

  q:=0;

  for i:=0 to m-1 do
    begin
      if BinPow(i,n)=y then
      begin
        inc(q);
        ans[q]:=i;
      end;
    end;

  for i:=1 to q do
    write(ans[i],' ');

  if q=0 then writeln(-1);
end.

#include <bits/stdc++.h>

using namespace std;
string a,b;
string re(long long c, long long d,string x){
    string y="";
    for(long long i=d; i>=c; i--) y=y+x[i];
    return y;
}
string re2(string x){
    long long t=0;
    string res="";
    for(long long i=0; i<x.size(); i++){
        int q=(int)x[i];
        if(q<65||(q>90&&q<97)||q>122){
            res=res+re(t,i-1,x);
            res=res+x[i];
            t=i+1;
        }
    }
    res=res+re(t,x.size()-1,x);
    return res;
}
int main()
{
    while(getline(cin,a)){
        istringstream iss;
        iss.clear();
        iss.str(a);
        while(iss>>b) cout<<re2(b)<<" ";
        cout<<endl;
    }
}

"King Hugo II wants to state the number x in his recruitment order. The king needs as many soldiers as possible for his military campaign. However, if the recruitment takes more than t days, the enemy may learn about the imminent intrusion and strike first. Help Hugo II find the maximal possible value of x."

We must find "maximal possible value of x" with "if the recruitment takes more than t days, the enemy may learn about the imminent intrusion and strike first" without "The king needs as many soldiers as possible for his military campaign"!!!

nvm accepted

Edited by author 05.02.2017 04:33