如何使用大数组大小

我在spoj上尝试这个问题。www.spoj.com/problems/RRANGE.It需要段树。但问题在于数组的大小。这里是（1<=N<=1000000）。有办法解决这个问题吗？这是我的实现（给出了接近N<1000000的正确答案）

#include <stdio.h>
#include <math.h>
#include<iostream>
#include<string.h>
using namespace std;
//segment tree
long long a[1000000];
long long Mid(long long s,long long e)
{
    return s+(e-s)/2;
}
long long Sum1(long long *st,long long ss,long long se,long long qs,long long qe,long long index)
{
    if (qs<=ss&&qe>=se)
        return st[index];
    if (se<qs||ss>qe)
        return 0;
    long long mid=Mid(ss, se);
    return Sum1(st,ss,mid,qs,qe,2*index+1) +Sum1(st,mid+1,se,qs,qe,2*index+2);
}
void update1(long long *st,long long ss,long long se,long long i,long long diff,long long index)
{
    if (i<ss||i>se)
        return;
    st[index]=st[index]+diff;
    if (se!=ss)
    {
        long long mid = Mid(ss,se);
        update1(st,ss,mid,i,diff,2*index+1);
        update1(st,mid+1,se,i,diff,2*index+2);
    }
}
void update(long long arr[],long long *st,long long n,long long i,long long new_val)
{
    if (i<0||i>n-1)
    return;
    long long diff = new_val - arr[i];
    arr[i] = new_val;
    update1(st,0,n-1,i,diff,0);
}
long long Sum(long long *st,long long n,long long qs,long long qe)
{
    if (qs<0||qe>n-1||qs>qe)
    return -1;
    return Sum1(st,0,n-1,qs,qe,0);
}
long long segtree(long long arr[],long long ss,long long se,long long *st,long long si)
{
    if (ss==se)
    {
        st[si]=arr[ss];
        return arr[ss];
    }

    long long mid=Mid(ss, se);
    st[si]=segtree(arr,ss,mid,st,si*2+1)+segtree(arr,mid+1,se,st,si*2+2);
    return st[si];
}
long long *segt(long long arr[],long long n)
{
    long long x = (long long)(ceil(log2(n)));
    long long max_size = 2*(long long)pow(2, x) - 1;
    long long *st = new long long[max_size];
    segtree(arr,0,n-1,st,0);
    return st;
}
int main()
{
    //memset(a,0,sizeof(a));
    long long n,u,v;
    cin>>n>>u>>v;
    for(long long i=0;i<n;i++)
    a[i]=0;
    long long *st=segt(a,n);

    while(u--)
    {
        long long i,j;
        cin>>i>>j;
        long long z=1;
        for(long long p=i-1;p<j;p++)
        {
        update(a,st,n,p,a[p]+z);
        z++;
        }
    //for(int m=0;m<n;m++)
    //cout<<a[m]<<endl;
    }
    while(v--)
    {
        long long i,j;
        cin>>i>>j;
        cout<<Sum(st,n,i-1,j-1)<<endl;
    }
    return 0;
}