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Discussion of Problem 1005. Stone Pile

solution
Posted by Md.Forhad Hossain 29 Jan 2011 01:55
#include<iostream>
using namespace std;
int main(void)
{
    int input,weight,temp,i;
    long int sum=0;
    cin>>input;
    if(input<1 && input>21)
        exit(0);
    int ary[20];
    for(i=0;i<input;i++)
    {
        cin>>weight;
        if(weight>=1 && weight<=100000)
        {
            ary[i]=weight;
            sum+=weight;
        }
    }
    for(i=0;i<input-1;i++)
    {
        for(int j=0;j<input-i-1;j++)
        {
            if(ary[j]>ary[j+1])
            {
                temp=ary[j];
                ary[j]=ary[j+1];
                ary[j+1]=temp;
            }
        }
    }
    int x=input-1;
    if(x==0)
    {
        cout<<ary[x];
        exit(0);
    }
    long int k1,k2,middle;
    middle=sum/2;
    k2=ary[x];
    k1=ary[x-1];
    int r=0;
    int bol=0;
    for(i=input-3;i>=r;)
    {
        if((k1+k2)<=middle)
        {
            k2=k1+k2;
            k1=ary[i];
            i--;
        }
        else if(k1<k2)
        {
            if(bol==0)
            {
            k1+=ary[r];
            r++;
            bol=1;
            }
            else
            {
                k1+=ary[i];
                bol=0;
                i--;
            }
        }
        else
        {
            if(bol==0)
            {
            k2+=ary[r];
            r++;
            bol=1;
            }
            else
            {
                k2+=ary[i];
                bol=0;
                i--;
            }
        }
    }
    if(k2>k1)
        cout<<k2-k1;
    else
        cout<<k1-k2;
    return 0;
}
Re: solution
Posted by tanu 20 Mar 2011 13:19
can u pls explain ths code..atr sortng wht hv u dn??
Re: solution
Posted by tanu 20 Mar 2011 13:20
can u pls explain ths code..aftr sortng wht hv u dn??
Re: solution
Posted by ScaV 29 Aug 2011 17:36
 if(input<1 && input>21)
??? is it possible?
Re: solution
Posted by zdw1224 24 Nov 2011 21:34
Consider the following situation : 3221

1) k2=3,k1=2
2) k1+k2=5>4 bol=0 k2>k1, so k1=k1+1=3, k2=3, bol=1
3) k1+k2=6>4 bol=1 k1>=k2, so k2=k2+2=5, k1=3, bol=0

-> result = 2,   but   31 | 22  -> 0

!!!