Monday 11 January 2016

Solution of LIGHT OJ 1227 :: Boiled Eggs


Before seeing the solution make sure that you tried enough. Don’t paste the whole code, just find out the logic. If you stuck in trouble, just inform me on comment.

/**Bismillahir Rahmanir Rahim.**/

#include <stdio.h>
int main()
{
    int n, p, q, tst, i, j;
    scanf("%d", &tst);
    for(i=1; i<=tst; i++)
    {
        scanf("%d%d%d", &n, &p, &q);
        int ara[n], cnt=0, sum=0;
        for(j=0; j<n ;j++)
        {
            scanf("%d", &ara[j]);
            sum += ara[j];
            if(cnt<p && sum<=q) cnt++;
        }
        printf("Case %d: %d\n", i, cnt);
    }
    return 0;
}

1 comment:

  1. #include
    using namespace std;
    int main()
    {
    int t , j , count;
    while(1==scanf("%d",&t))
    {
    j=0;
    while(t--)
    {



    int n , p , q ;
    cin>>n>>p>>q;
    int i = 0 , arr[n];
    for(i=0;i>arr[i];
    }
    std::sort(arr,arr+n);
    if(n=0 && arr[i]>0)count++;
    else break;
    i++;
    }
    cout<<"Case "<<++j<<": "<0)
    count++;

    }
    if(sum>q)i--;



    cout<<"Case "<<++j<<": "<<count<<endl;

    }



    }

    }


    }



    please help me by finding out a test case that may give wrong ans of this problem. i couldnt find such case. but im having wrong ans frequently .

    ReplyDelete

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