del

Edited by author 06.01.2015 21:37

 Test #01 куда тут ошибка помогите пожалуйста
#include <iostream>
using namespace std;
int main ()
{
    short int A[6];
    signed  int s=0,s2=0;
    for (int i=0;i<6;i++)
    cin>>A[i];
    for (int i=0;i<3;i++)
     s=s+A[i];
    for (int i=5;i>=3;i--)
     s2=s2+A[i];

     if ((s-s2==-1)  || (s-s2==1 && A[5]!=9))
     cout<<"Yes";

     else
     cout<<"No";
    //system ("pause");

    return 0;

    }

Edited by author 22.06.2012 19:44

Edited by author 21.06.2014 02:14

var a,b,c,d,e,f:integer;
begin
 readln(a,b,c,d,e,f);
if ((a+b+c)-(d+e+f))=1  then writeln ('Yes') else
if ((d+e+f)-(a+b+c))=1 then
writeln ('Yes') else writeln ('No');
end.

#include <iostream>
using namespace std;

int main()
{
    int a,b,c;
    cin >> a;
    b=a-1;
    c=a+1;
    if (a%10+(a/10)%10+(a/100)%10==(a/1000)%10+(a/10000)%10+(a/100000)%10 || b%10+(b/10)%10+(b/100)%10==(b/1000)%10+(b/10000)%10+(b/100000)%10 || c%10+(c/10)%10+(c/100)%10==(c/1000)%10+(c/10000)%10+(c/100000)%10)
        cout << "YES";
    else
    {
        cout << "NO";
    }
    return 0;
}

What in test #8?

Edited by author 25.07.2013 00:58

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

 int main()
{
 int arr[5],arr1[5],arr2[5],i,a=0,b=0,n,x=0,y=0;
 printf("Enter number");
 scanf("%d",&n);
 x=n+1;
 y=n-1;
 while(n>0)
 {
    for(i=5;i>=0;i--)
    {
      arr[i]=n%10;
      n=n/10;
    }
 }


 a=( arr[0]+arr[1]+arr[2]);
 b=(arr[3]+arr[4]+arr[5]);
 //  printf("%d%d",a,b);
   if((a-b==1)||(a-b==-1))
      {
           while(x>0)
         {
           for(i=5;i>=0;i--)
           {
               arr1[i]=x%10;
               x=x/10;
           }
         }
          while(y>0)
         {
           for(i=5;i>=0;i--)
           {
               arr2[i]=y%10;
               y=y/10;
           }
         }
       if ((arr1[0]+arr1[1]+arr1[2]==arr1[3]+arr1[4]+arr1[5])||(arr2[0]+arr2[1]+arr2[2]==arr2[3]+arr2[4]+arr2[5]))
       { printf("\nYes");}
        else
       { printf("\nNo");}
      }
      else
      {
          printf("No");
      }
    return 0;
}

Please, give me some tests, i don't know, what wrong in my program(

Here is my code. Give me some example to test:
def sumnum(n):
    n=str(n)
    a=0
    for i in n:
        a+=int(i)
    return a
a=int(input())
num=[a+1,a-1]
l=1
b=''
c=''
for a in num:
    a=(str(a))
    b=a[:len((a))//2]
    c=a[len((a))//2:]

    b=sumnum(b)
    c=sumnum(c)
    if b==c:
        print ('Yes')
        l=0
        break
if l:
    print ('No')

main()
{
long n,sumperv=0,sumposled=0,perv,posled;
int flag=0;

scanf ("%ld",&n);
perv=(n+1)/1000;
posled=(n+1)%1000;
while (perv) {
      sumperv+=perv%10;
      perv/=10;
}
while (posled) {
      sumposled+=posled%10;
      posled/=10;
}
if (sumperv==sumposled)
   flag=1;

perv=(n-1)/1000;
posled=(n-1)%1000;
sumperv=0; sumposled=0;
while (perv) {
      sumperv+=perv%10;
      perv/=10;
}
while (posled) {
      sumposled+=posled%10;
      posled/=10;
}
if (sumperv==sumposled)
   flag=1;

if (flag==1)
   printf ("yes");
else
    printf ("no");



}

var a:array[1..6]of integer;
    i:integer;
    g:char;
    s:string;
begin
read(s);
for i:=1 to 6 do
val(s[i],a[i],g);
if (a[6]<>9) and
((a[1]+a[2]+a[3]=a[4]+a[5]+a[6]+1)or
(a[1]+a[2]+a[3]=a[4]+a[5]+a[6]-1))then
writeln('Yes') else writeln('No');
end.

"either the previous or the next ticket is lucky." and "one step from happiness"

lucky means: previous OR next, sum of first three digits EQUAL sum of last three digits.

Edited by author 03.10.2012 20:54

715068 Yes          7+1+5=13  0+6+8=14
445219 No           4+4+5=13  2+1+9=12     <<<---- Why ANSWer is No ? ? ?
012200 Yes          0+1+2=3   2+0+0=2

WTF ?

Edited by author 25.07.2011 00:19

Edited by author 25.07.2011 02:11

I couldn't understand the statements.

Now, I could be saying something stupid here.  But I think we're missing a definition of what happens at the topmost boundary.
Say our number is 999,999, if you subtract one, you don't get a lucky number, but if you add one, then what happens?
The problem said that it was going to be a 6 digit number at all times, but if you add one, it becomes a seven digit number, so does this case never occur?  And if it does, then does the lucky number formula apply only to the last 6 digits of the number?

 why a programe crash?

either the previous or the next ticket is lucky ---- > what does the lucky means ? can anyone give me an example?

#include<iostream.h>
#include<math.h>
#include<stdlib.h>
bool prov(char s[10])
{
int k1=0,k2=0;

k1=s[0]-'0'+s[1]-'0'+s[2]-'0';

k2=s[3]-'0'+s[4]-'0'+s[5]-'0';

if(abs(k2-k1)==1)return true;else return false;

}
int main()
{
char s[10],str1[10];
int pr=0,k1,k2;
cin>>s;
if(prov(s)==true)
{
ltoa(atol(s)+1,str1,10);
k1=str1[0]-'0'+str1[1]-'0'+str1[2]-'0';
k2=str1[3]-'0'+str1[4]-'0'+str1[5]-'0';
if(k1==k2)pr=1;

ltoa(atol(s)-1,str1,10);
k1=str1[0]-'0'+str1[1]-'0'+str1[2]-'0';
k2=str1[3]-'0'+str1[4]-'0'+str1[5]-'0';
if(k1==k2)pr=1;

if(pr==0)cout<<"No";else cout<<"Yes";
}else cout<<"No";

return 0;


}

this is my code, but wa in test 1, why?
what is the test 1,
thanks
sorry my english is very poor
    import java.io.*;
    import java.util.*;

    public class Main{


        static int sum(int n)
        {
            int sum=n%10+(n/10)%10+(n/100)%10;
            return sum;
        }

        static boolean solve(String n)
        {
            String ant=n.substring(0,3);
            String post=n.substring(3,6);
            if(ant.compareTo(post)==0)
               return true;
            else
            {

            int antn=Integer.parseInt(ant);
            int postn=Integer.parseInt(post);

            int a;
            int b;
            int c;
            int d;
            if(postn+1>999)
            {
                a=antn+1;
                b=0;
            }
            else
            {
                a=antn;
                b=postn+1;
            }
            if(postn-1<0)
            {
                c=antn-1;
                d=999;
            }
            else
            {
                c=antn;
                d=postn-1;
            }
            a=sum(a);
            b=sum(b);
            c=sum(c);
            d=sum(d);

            if(a==b||c==d)
               return true;
            else
             return false;
            }
        }


        public static void main (String args[])
        {
            Scanner in=new Scanner(System.in);
            String n;
            while(in.hasNext())
            {
              n=in.next();
              if(solve(n))
                System.out.println("YES");
              else
                System.out.println("NO");
            }
        }
    }