#include <iostream>

using namespace std;
bool prime(long long n);


int main()
{
    long long a,odd, i, j, nice,b;
    cin >> a >> b;
    nice=0;
    if (a%2==1) {odd=a;} else {odd=a+1;}
    for (i=b; i>a; i--)
    {
       if (prime(i))
       {
           nice=i;
           break;
       }
    }

    if ((a==1 or nice==0) and b!=a) { cout << odd;}
    else if (b==a) {cout << a;}
    else
    if (nice) {cout << nice << endl;}
    return 0;
}

bool prime(long long n)
{
    bool w;
    w=n==2;
    if (!w and n%2)
    {
        w=1;
        for (int i=3; i*i<=n; i+=2)
        {
            if (n%i==0)
            {
                w=0 ;
                break;
            }
        }
    }
    if (w)
    {
        return 1;
    }
    else return 0;
}

2 4
2 6
2 8
4 2
4 4
4 6
4 8

Please somebody help me. I can not find the reason not to be accepted my code.

#include <iostream>
#include <string>
#include <vector>
using namespace std;
bool found;
string convert(string str)
{
    string char_list = "22233344115566070778889990";
    string to_digit;
    for(int i=0; i<str.length(); i++)
    {
        to_digit += char_list[str[i]-97];
    }
    return to_digit;
}
void dfs(vector<int> index[], vector<int> ends[], string diction[], vector<int> track, int i, int n)
{
    if(found == true)
    {
        return;
    }
    if(i == n){
        for(int j=0; j<track.size(); j++)
        {
            cout<<diction[track[j]]<<" ";
        }
        found = true;
        return;
    }
    for(int k=0; k<ends[i].size(); k++){
        track.push_back(index[i][k]);
        dfs(index, ends, diction, track, ends[i][k], n);
        track.pop_back();
    }
}
int main()
{
    for(; ;){
        found = false;
        string number;
        cin>>number;
        if(number == "-1"){
            break;
        }
        int dic_size;
        cin>>dic_size;
        string diction[dic_size];
        for(int i=0; i<dic_size; i++){
            cin>>diction[i];
        }
        int n = number.size();
        vector<int> ends[n];
        vector<int> index[n];
        for(int j=0; j<dic_size; j++)
        {
            string str = convert(diction[j]);
            int len = diction[j].size();
            for(int i=0; i<n; i++)
            {
                if(number.substr(i,len) == str){
                    ends[i].push_back(i+len);
                    index[i].push_back(j);
                }
            }
        }
        vector<int> track;
        dfs(index, ends, diction, track, 0, n);
        if(found == false){
            cout<<"No solution.";
        }
    }
    return 0;
}

Edited by author 03.01.2017 20:24

My algo is O(N*N*sqrt(N)).

var g:array[1..25000,1..25000] of integer; v,i,j,n,f,un,uk,t,k,s,x:longint;
Q:array[1..100000] of integer;
p:array[1..25000] of integer;
begin
assign(input,'input.txt');
assign(output,'output.txt');
reset(input);rewrite(output);
readln(t);
for i:=1 to t do
begin
  readln(n);k:=0;
  fillchar(g,sizeof(g),0);
  for j:=1 to n do begin read(s);
                         if s=0 then begin f:=j; k:=k+1;g[j,j]:=0;end
                                else begin  g[j,s]:=1;g[s,j]:=1;end;
                         if k>1 then break;
                   end;
  readln;
  if k<>1 then begin Writeln('NO');end else
  begin
    for j:=1 to n do p[j]:=-1;
    fillchar(Q,sizeof(Q),0);
    un:=1;
    uk:=2;
    Q[1]:=f;
    p[f]:=0;
    x:=0;
    while un<>uk do
    begin
      v:=Q[un]; un:=un+1;
      for j:=1 to n do
      if (g[v,j]<>0) and (p[j]=-1) then begin
                                        Q[uk]:=j;
                                        uk:=uk+1;
                                        p[j]:=p[v]+1;
                                        end;

    end;
    for j:=1 to n do
      if p[j]=-1 then begin x:=1;break;end;
      if x=0 then writeln('YES') else writeln('NO');
  end;
end;
end.

Could anyone give me some tests for this one?

Could someone tell me what's the meaning of this problem?my english is noot good.The google translation is bad.

Split your land and get smaller lands !

First, you have a land (1,1)(n,n)

For example:

Your land:

1 1 1 1 1 1
1 1 1 1 1 1
1 1 1 1 1 1
1 1 1 1 1 1
1 1 1 1 1 1
1 1 1 1 1 1

after adding the land with maximal importance:

1 1 1 1 1 1
1 1 1 1 1 1
1 1 9 9 1 1
1 1 9 9 1 1
1 1 1 1 1 1
1 1 1 1 1 1

You have 4 new lands:

2 lands:

a a a a a a
a a a a a a
1 1 9 9 1 1
1 1 9 9 1 1
a a a a a a
a a a a a a

and 2 lands:

a a 1 1 a a
a a 1 1 a a
a a 9 9 a a
a a 9 9 a a
a a 1 1 a a
a a 1 1 a a

For each of your new lands, you continue add other lands with decreasing importance.

Finally, you will find out that after adding a land with a certain importance t, you can't create a land with desired park size. Now, you end your program and output t.

THE END ^^!







Edited by author 08.10.2010 14:39

Edited by author 12.10.2010 10:14

in  more languages  "3."  or "0.E+6"  numbers are legal.
but for there are not, i.e.
input:
3.
1
0.E6
3
#
output:
Not a floating point number
Not a floating point number

Edited by author 02.01.2017 21:42

Could someone give me some tests for this one?

Try this test case:

1
1 of vodka
1
2 of vodka

Answer = 1

Could anyone, please, explain it to me? I can't understand how to program it. I'm filling the "amount of name" for the storage in the shop and for the queue, then, for each human from the queue, I check, firstly, whether the shop has the product the customer needs to buy, and then, if the shop has it, I compare the amounts. But I have a problem in understanding the part where the shop doesn't have enough of the product and the customers switch places. Could anyone explain it to me?

#include <stdio.h>
int main()
{
    int N, M;
    scanf("%d%d",&N,&M);
    long int res;
    if (N <= M)
        res = 2 * (N - 1);
    else
        res= 2 * (M - 1)+1;
    printf("\n%d",res);
    return 0;
}

where is the problem
#include<stdio.h>

int main(void)
{
 int a,b,sum;
 printf("a:");
 scanf("%d",a);

 printf("b:");
 scanf("%d",&b);

 sum=a+b;
 printf("Sum=%d",sum);

 return 0;
}

Edited by author 03.12.2015 01:58

I really can't solve this task?
Alwaus got TLE on test 39 :-(
Can someone give me hint or ...

I had a program like that:
while (x <= b/2) {
   ....
   x += dx
}
but it will fail test 1 1 100 (the last piece won't show).
Changed to
while (x <= b/2 + 0.000001) {
   ....
   x += dx
}

Got AC!

I Solved this problem, by transforming x and y into every system and search for the longest common substring.
But I am interested in another Ideas, better than mine, except brute force.

Нас просят найти количество непустНых палиндромов.

whats test #1??