RPLN / Negative Score

[equation_sof]

some one please help me my code is giving SIGSEGV runtime error!!
i hv deleted my alocated dymanic merory location also...
here is my code .....

#include<iostream>
#include<algorithm>
#include<cmath>
#include<string.h>
using namespace std;
void segment_tree(int node,int l,int r,int *a,int *arr){
     if(l==r)
     a[node]=l;
     else{
          int  mid=(l+r)/2;
          segment_tree(2*node,l,mid,a,arr);
          segment_tree(2*node+1,mid+1,r,a,arr);
          if(arr[a[2*node]]<=arr[a[2*node+1]])
          a[node]=a[2*node];
          else
          a[node]=a[2*node+1];
          }
          }
int query(int node,int start,int end,int *a,int *arr,int i,int j){
    if(i>end || j<start)
    return -1;
    if(i<=start && j>=end)
    return a[node];
    int mid=(start+end)/2;
    int p1=query(2*node,start,mid,a,arr,i,j);
    int p2=query(2*node+1,mid+1,end,a,arr,i,j);
    if(p1==-1)
    return p2;
    if(p2==-1)
    return p1;
    if(arr[p1]<=arr[p2])
    return p1;
    else
    return p2;                                             
             }
int main(){
    int t,cs=0;
    scanf("%d",&t);
    while(t--){
               int n,q,arr[1005];
               scanf("%d%d",&n,&q);
               for(int i=0;i<n;i++)
               scanf("%d",&arr[i]);
    //making a new array to store location           
    int x=(int)(ceil(log2(n)))+1;
    int size = 2*(int)pow(2.0,x);
    int *a=new int[size];
    memset(a,-1,sizeof(a));
    segment_tree(1,0,n-1,a,arr);
    /*for(int i=0;i<size;i++)
    cout<<a[i]<<" ";*/
    printf("Scenario #%d:\n\n",++cs);
    while(q--){
             int i,j;
             scanf("%d%d",&i,&j);
             int loc=query(1,0,n-1,a,arr,i-1,j-1);
             printf("%d\n",arr[loc]);
             }
    delete[] a;
    //end of one test case         
    }
    }

[sarora111]

hare is my code getting wrong ans at 9th test case.
@leppy sir:
Is there any presion error in calculting sqrtl(N)??
pls help.

#include<iostream>
#include<stdio.h>
#include<vector>
#include<cmath>
#include<algorithm>
using namespace std;
int main()
{
    int t;
    scanf("%d",&t);
    for(int T=1;T<=t;T++)
    {
      int N,Q;
      scanf("%d%d",&N,&Q);
      long long int score[N];
      int var,size,each;
      var=size=llrintl(sqrtl(N));
      if(var*var ==N)
      var=var;        
      else
      var=var+1;
      int M[var];
      int k=1;
      if(N%var==0)
      each=N/var;
      else
      each=N/var+1;
      long long int minimum=2000000000,count=0,len;
      for(int i=1;i<=N;i++)
      {
          scanf("%lld",&score[i]);
          count++;
          if(score[i]<minimum)
          {
             minimum=score[i];
             M[k]=i;
          }
          if(count==each)
          {
            count=0;
            minimum=2000000000;
            k++;
          } 
      }
      /*for(int i=1;i<=var;i++)
      cout<<M[i]<<" ";*/
      int A,B;
      printf("Scenario #%d:\n\n",T);
      for(int j=1;j<=Q;j++)
      {
      scanf("%d%d",&A,&B);
      int Amod,Adiv,Bmod,Bdiv;
      Bdiv=B/each;
      Bmod=B%each;
      Amod=(A-1)%each;
      Adiv=A/each;
      minimum=2000000000;
      for(int i=B;i>Bdiv*each;i--)
      minimum=min(minimum,score[i]);
      if(Amod==0)
      {
       for(int i=Adiv+1;i<=Bdiv;i++)
       minimum=min(minimum,score[M[i]]);
      }
      else
      {
         int save;
         //cout<<"abc\n";
         for(int i=A;;i++)
         {
           minimum=min(minimum,score[i]);
           if(i%each==0)
           {
              save=i;
              break;                   
           }
         }
         save=save/each;
         for(int i=save+1;i<=Bdiv;i++)
         minimum=min(minimum,score[M[i]]);
      }
      printf("%lld\n",minimum);
      }
    }
    //system("pause");
    return 0;
}

[leppyr64]

I don't understand yor algorithm.

[sarora111]

I used the logic given at the community.topcoder.com/tc?module ... onAncestor
RMQ

[leppyr64]

That's a cool algorithm. I learned something today

score[N] is undefined. You're also not implementing the algorithm correctly.