who can help me?
I use bfs.

Please, give me test 12. Very need.

Don't forget about it ;)

If WA8, try test
100000
PUSH 1 1
PUSH 1 2
....
PUSH 1 99990
POP 1
...
POP 1
Right answer:
99990
...
99981

What answer for
3
1
0
3
0
?

0
0
0

or

0

?

give me test plz

import java.util.Scanner;

public class Main {

    public void solucion(int num) {

        int aux = 0;
        if (num <0) {
            for (int i = num; i <= 1; i++) {
                aux += i;
            }
        } else {
            for (int j = num; j >= 1; j--) {
                aux += j;
            }
        }
        System.out.println(aux);

    }

    public static void main(String[] args) {
        Main obj = new Main();
        Scanner leer = new Scanner(System.in);
        int numero = leer.nextInt();
        if (numero < 10000) {
            obj.solucion(numero);
        }


    }
}

#include <iostream.h>
int main(){
    int num;
    while (cin>>num){
          int answer=0;
          if(num>=1){
                     answer = num*(num+1)/2;
                     printf("%d\n",answer);
                     }
          else{
               answer = (2-num)*(1+num)/2;
               printf("%d\n",answer);
               }
          }
          return 0;
    }
where ?

what`s wrong?


var n,i,res,c,max:longint;
a:array[0..100000] of integer;
begin
repeat
max:=0;
Readln(n);
a[0]:=0;
a[1]:=1;
for i:=1 to n do
begin
a[2*i]:=a[i];
a[2*i+1]:=a[i]+a[i+1]
end;
for i:=1 to n do
if a[i]>max then max:=a[i];
if n<>0 then Writeln(max);
until n=0;
end.

Subject

This code gets WA on test 2:
[code deleted]

& this gets AC:
[code deleted]

:)

Edited by moderator 02.01.2020 18:26

I know O(n^2) one. Maybe there is also an O(1)?

#include <stdio.h>

int main(void) {
int n,k,i;
char c;
scanf("%d",&n);
while(n--){
scanf("%1s%d",&c,&k);
if((c=='a'||c=='h')&&(k==1||k==8)) {printf("2\n"); continue;}
if((c=='a'||c=='h')&&(k==2||k==7)) {printf("3\n"); continue;}
if((c=='b'||c=='g')&&(k==1||k==8)) {printf("3\n"); continue;}
if((c=='b'||c=='g')&&(k==2||k==7)) {printf("4\n"); continue;}
if(c<'b'||c>'g') {printf("4\n"); continue;}
if(k<2||k>7) {printf("4\n"); continue;}
if((c=='b'||c=='g') || (k==2||k==7)) {printf("6\n"); continue;}
printf("8\n");
}
return 0;
}

            long stringLength = Convert.ToInt32(Console.ReadLine());
            string S = Console.ReadLine();
            string T = Console.ReadLine();
            long temp = 0;
            bool check = false;
            for (long i = 0; i < stringLength; i++)
            {
                S = S[Convert.ToInt32(stringLength) - 1] +
                                     S.Substring(0,  Convert.ToInt32(stringLength) - 1);
                if (S == T)
                {
                    temp = i + 1;
                    check = true;
                    break;
                }
            }
            if (check && temp < stringLength)
                Console.WriteLine(temp);
            else
                Console.WriteLine(-1);

The printing of the answer in C++ should be, e.g., in the form:
cout << "Distance is "
     << setiosflags(std::ios_base::fixed)
     << setprecision(0)
     << 1000.0*sqrt(dist) << " km.";
where dist is double value. You don't need to round off the answer with the help of some functions.

Please give a test № 1 for the task № 1601

Why this solution got AC ? For the first test it outputs 15, but correct answer is 3 !!!!
#include <iostream>
#include <complex>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
using namespace std;
int main()
{
    int n,i,total=0,counter;
    int l;
    vector <int> x;
    cin>>n;
    for(i=0; i<n; i++)
    {
        cin>>l;
        x.push_back(l);
    }
    for(i=0; i<n; i++)
    {
        total+=x[i];
    }
    counter=total/2;
    for(i=0; i<n; i++)
    {
        if(x[i]<=counter)
        {
            counter-=x[i];
        }
    }
    cout<<2*counter+total%2;
    return 0;
}

I have tested my program on many tests, and I don't see any mistakes in my algorithm. What does mean "Output limit exceeded"? Please help. Thanks.

Hint for you :

int travel(int player,int node)
{
    int value;

    if(isLeaf(node)) return getValue(node);

    // If the first player is at this node, he prefer getting the maximal value among -1, 0 and +1.
    if(player == 0)
    {
        value = MinValue();
        for(int a=0;a<numberOfChild(node);a++)
        {
            int v = travel((player+1)%2,getChild(node,a));
            if(value < v) value = v;
        }
    }
    // Get the minimal value.
    else
    {
        value = MaxValue();
        for(int a=0;a<numberOfChild(node);a++)
        {
            int v = travel((player+1)%2,getChild(node,a));
            if(value > v) value = v;
        }
    }
    return value;
}

Edited by author 24.02.2011 15:14

#include <cstdio>
using namespace std;
 int main()
{
     int const Q=1001;
     long long unsigned k = 0,i,n,a=0;
    scanf("%llu",&n);
    long long unsigned mas1[Q];
    long long unsigned mas2[Q];
    for(i = 1;i<=n;i++)
    { scanf("%llu",&mas1[i]);
     mas2[i] = mas1[i];
    }
    for(;;)
   {
       for(i = 1;i<=n;i++)
     {
         if (i == mas2[i]){ k++;}
         mas2[i] = mas1[mas2[i]];
     }
      a++;
     if ( k == n) { break;} else {k = 0;}
   }
    printf("%llu\n",a);
   return 0;
}