i have WA #2
#include <iostream>
#include <stdio.h>
using namespace std;

int main()
{
    int n, m;
    cin >> n >> m;

    int win = 0;
    for(int i = 0; i < m; i++)
    {
        int a;
        cin >> a;
        n -= a;
        if(i % 2 == 0)
            win = 2;
        else win = 1;
    }

    cout << win << endl;

    return 0;
}

I think, this problem should have lower difficulty...

15 3
1 5 11
correct output:1

This is a big point I think should take care.

what the algo of this problem? please^_^

4 3 1 1 3 1 4 or 1 3 4 1 3 1 4
sum of all is 17
and first player wins not second,
can somebody explan?


Edited by author 18.10.2012 17:08

Edited by author 18.10.2012 17:11

can it be solved with backtracking?
but i think recursion depth will exceed memory and also lime will not pass, right?

Can someone write the sequence of stone that lead to player 2 lose plz ?

Sample? please

15 3
3 5 7

AC return "1", but the right answer is "2", isn't it ??

I think, that tests for this problem are bad.
For test 3 3 3 2 1 my AC program gives answer 2, while answer 1 is correct.
P.S. My first AC solution is incorrect, but i hope, that my second AC solution is correct.

a lot of thnx

#include<math.h>
#include<stdio.h>


int main()
{
    int t[100],n,m,i,j,tr[10010];
    scanf("%d",&n);
    scanf("%d",&m);
    for(i=1;i<=m;i++)
    {

                    scanf("%d",&t[i]);
    }
   for(i=1;i<=n;i++)
    {
                    if(i<=t[1])
                        tr[i]=2;
                    else
                    {
                        for(j=1;j<=m;j++)
                        {
                             if(i>t[j])
                             {printf("hhhhhhhhhhhh %d  %d  %d \n",i,t[j], tr[i-t[j]]);
                                if(tr[i-t[j]]==2)
                                    {tr[i]=1;
                                    printf("hahaha\n");
                                    break;
                                    }
                             }
                        }
                        if(j==m+1)
                            tr[i]=2;

                    }
                    printf("%d   %d\n",i,tr[i]);
    }

    printf("%d",tr[n]);


return 0;
}

(excuesme for my bad english)
if the test like this:
5 3
2 4 5
correct ans is: 2
because:
stones:-------0-1-2-3-4-5
winner player:1-0-2-0-2-2 (0 for unknown position)
but AC solution returns 1

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

void main()
{
    int i,j,len,count=0,m,n;
    char ch[2001];
    scanf("%s",ch);
    len=strlen(ch);
    if(len%2==0)
    {
       m=i=len/2-1;
        n=j=len/2;
    }
    else
    {
        m=i=len/2;
        n=j=len/2;
    }
    for(m,n;m>=0;m--,n++)
    {
        if(ch[m]>ch[n])
        {
           break;
        }
        if(ch[m]<ch[n])
        {
           count=1;
           break;
        }
    }
    if(count)
    {
        ch[i]=ch[i]+1;
    }
    else
    {
       ch[i]=ch[i];
    }
    for(;i>=0;i--,j++)
    {
        ch[j]=ch[i];
    }
    ch[j]='\0';
     printf("%s\n",ch);

}

what's the date?

Moya programma ne prohodit test#5... Na chto on napravlen???