4 100
aaaaaaaaaaaaaaaaaaaa11aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa111
ans : 4

Edited by author 15.10.2006 11:24

346 158 618
955.0 105.0
891.0 93.0

65.1152, isn't it?????

Edited by author 21.04.2008 16:38

#include <iostream>
using namespace std;
int main()
{
    int n;
    cin>>n;
    if(1<=n&&n<=12)
    {
        int a=12-n;
    if(a*45<=4*60)
    cout<<"YES";
    else
    cout<<"NO";
    }
return 0;
}

Tell me please how you have get timing 0.001 ? It mustn't be a code, may be just method or algorithm. Thanks

If I divide first and then add 1 with (n or other group numbers), I get WA. But if I add 1 at first and then I divide that time i get AC.
I have seen some of the discussion, but couldn't find the logic.Can someone help me out?

#include <iostream>
#include <iomanip>
#include <map>
#include <vector>
#include <cmath>
#include <vector>
#include <algorithm>
#include <cmath>

using namespace std;

int letter(char c)
{
    return (('a'<=c&&c<='z')||('A'<=c&&c<='Z'));
}

int main()
{
    //freopen("a","r",stdin);
    string s;
    while(getline(cin,s))
    {
        vector<int> l,r;
        for (size_t i=0;i<s.length();i++)
        {
            if (letter(s[i]))
            {
                if (!i||!letter(s[i-1]))
                {
                    l.push_back(i);
                }
                if (i+1==s.length()||!letter(s[i+1]))
                {
                    r.push_back(i+1);
                }
            }
        }
        for (int i=0;i<l.size();i++)
        {
            reverse(s.begin()+l[i],s.begin()+r[i]);
        }
        cout << s << endl;
    }
    return 0;
}

Edited by author 28.10.2017 19:36

I test all case in discussion,but all answers right...
Can someone help..Thanks!!!

I know how to construct larger cases when knowing solution n<=11&&m<=11,but brute_force for small tests is tooooooooo slow..

d = {'._.\n|.|\n|_|': 0,
     '...\n..|\n..|': 1,
     '._.\n._|\n|_.': 2,
     '._.\n._|\n._|': 3,
     '...\n|_|\n..|': 4,
     '._.\n|_.\n._|': 5,
     '._.\n|_.\n|_|': 6,
     '._.\n..|\n..|': 7,
     '._.\n|_|\n|_|': 8,
     '._.\n|_|\n._|': 9
     }

#include<stdio.h>
#include<math.h>
int main()
{
    int i,sum=0,n;
    scanf("%d",&n);

    if(abs(n)<=10000){

    if(n<0){

    for(i=1;i>=n;i--){
        sum+=i;
    }
    }

    else{
        sum=((n*n)+n)/2;
    }

    printf("%d",sum);}
    return 0;
}

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

int main()
{

    int a,p=0,i,*b;
float f,s=0;
scanf("%d",&a);
b=(int *)malloc(sizeof(int)*a);
for(i=0;i<a;i++){
scanf("%d",&b[i]);}
/*if (b[i]<3||b[i]>5)
    scanf("%d",&b[i]);
}*/
for(i=0;i<a;i++)
    s+=b[i];

f=s/a;
for(i=0;i<a;i++)
{

 if(b[i]==3){
 p=1;}
}

if(f==5)
   {printf("Named");
   }

       else if(f>=4.5&&p==0){ printf("High");
    }

        else if(f<4.5&&p==0){
    printf("Common");}
    else {printf("None");}
    free(b);
    return 0;
}

I've passed all of the user tests on here, and any test I can think of.
Still getting WA on #14.
Here's my non-working code.

import java.io.*;

public class P1998 {
    static StreamTokenizer in;
    static PrintWriter out;

    static int nextInt() throws IOException{
        do{
            in.nextToken();
        }while(in.ttype != StreamTokenizer.TT_NUMBER);
        return (int)in.nval;
    }

    public static void main(String[] args) throws IOException{
        in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        out = new PrintWriter(System.out);

        int n = nextInt();
        int m = nextInt();
        int k = nextInt();

        int[] stones = new int[n];
        int[] dropsTo = new int[n];

        int tmp = 0, j = 0;
        for(int i = 0; i < n; i++){
            stones[i] = nextInt();

            while(j < i - 1 && tmp - stones[j] > k){
                tmp -= stones[j];
                j++;
            }
            tmp += stones[i];
            dropsTo[i] = j;
        }

        j = 0;
        int moment, time = 0;
        for(int i = 0; i < m; i++){
            moment = nextInt() - time;

            time += moment;
            j += moment - 1;
            if(j >= n){
                time -= moment - n;
                j = n;
                break;
            }
            j = dropsTo[j];
        }

        out.println(time + n - j);
        out.flush();
    }
}

Create an array of Boolean elements. Size = billion.
For i:=1 to n {{ introduce a variable; element of array with index [variable] equals true }}
For i:=1 to m {{ introduce a variable; if element of array with index [variable] equals true then increase the counter }}

And finally, we output counter))
That's all), sorry for bad english

But this solution takes 58 mb of memory... ((

Edited by author 06.04.2014 16:17

Used Sieve to generate primes till 10^6, and then used those primes to generate larger primes.
counted all such numbers satisfying  L <= p^(q-1) <=R where p & q are primes (q>2). Subtracted it from R-L+1

How to optimize?

I got WA on test 4, but on all tests I could make myself and also on those, which were in the problem, my program answered correctly. Please, suggest some tests, that could help.

import java.util.Scanner;
import java.util.Stack;

public class ReverseRoot_SplitDecimal {

    public static void main(String args[]) {

        Scanner sc = new Scanner(System.in);
        Stack<Integer> integerNumberStack = new Stack<Integer>();
        Stack<Short> floatingNumberStack = new Stack<Short>();

        Double tempDouble = new Double(0);

        while(sc.hasNextInt()) {
            tempDouble = Math.sqrt(sc.nextDouble());
            integerNumberStack.push(Integer.parseInt(String.format("%.4f", tempDouble).split("\\.")[0]));
            floatingNumberStack.push(Short.parseShort(String.format("%.4f", tempDouble).split("\\.")[1]));
        }

        while(!integerNumberStack.isEmpty()) {
            System.out.printf("\n%d.%04d", integerNumberStack.pop(), floatingNumberStack.pop());
        }
        System.out.println();

        sc.close();
    }
}

Just drawing different patterns on paper for several hours and the solution will come!