Problem Link : http://www.spoj.com/problems/SEGSQRSS/
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This problem is a nice example of segment tree with lazy propagation. All you need to solve this kind of problem is to do some experiment with the formula and convert them into a suitable form. From which we can drive the segment tree solution.
In this problem given an integer array A=[a1,a2,a3,a4…]. You need to perform 3 kinds of operation on it.
type 0: Set all the element in the interval l to r to x.
type 1: Increase all the element in the interval l to r by x.
type 2: Print the value of square sum of the interval l to r. For example let l=1 and r=3. then you need to print a1^2 + a2^2 + a3^2.
At first think we have a pre-calculated square sum array. Then if we perform type 1 operation which is increase it by x then what scenario will happen?
Our sum now become (a1+x)^2 + (a2+x)^2 + (a3+x)^2. Let expand this equation-
(a1+x)^2 + (a2+x)^2 + (a3+x)^2
= a1^2+ 2*a1*x + x^2 + a2^2+ 2*a2*x + x^2 + a3^2+ 2*a3*x + x^2
= (a1^2 + a2^2 + a3^2) + 3*x^2 + 2*x*(a1 + a2 + a3)
for a range l to r the generalized equation is –
(a_l^2 + a_l+1^2 +….+ a_r^2) + (r-l+1)*x^2 + 2*x*(a_l + a_l+1 +….+ a_r)
We can store the information about sum, square_sum, type_0 lazy and type_1 lazy of an interval in each segment tree node. And perform type 1 operation then we can calculate then sum value for a node over a range by multiplication and square sum value by above equation.
For every type 0 operation we can simply put the value of sum and square sum by using some basic calculation and arithmetic multiplication operation.
And for every type 2 operation we just need to get the sum of square sum value over a range l to r.
So we can perform each of 3 operation using a segment tree with lazy propagation.
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define pii pair <int,int>
#define pll pair <long long,long long>
#define sc scanf
#define pf printf
#define Pi 2*acos(0.0)
#define ms(a,b) memset(a, b, sizeof(a))
#define pb(a) push_back(a)
#define MP make_pair
#define db double
#define ll long long
#define EPS 10E-10
#define ff first
#define ss second
#define sqr(x) (x)*(x)
#define D(x) cerr<<#x " = "<<(x)<<endl
#define VI vector <int>
#define DBG pf("Hi\n")
#define MOD 1000000007
#define CIN ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define SZ(a) (int)a.size()
#define sf(a) scanf("%d",&a)
#define sfl(a) scanf("%lld",&a)
#define sff(a,b) scanf("%d %d",&a,&b)
#define sffl(a,b) scanf("%lld %lld",&a,&b)
#define sfff(a,b,c) scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v) for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n) for(int i=0;i<n;i++)
#define loop1(i,n) for(int i=1;i<=n;i++)
#define REP(i,a,b) for(int i=a;i<b;i++)
#define RREP(i,a,b) for(int i=a;i>=b;i--)
#define TEST_CASE(t) for(int z=1;z<=t;z++)
#define PRINT_CASE printf("Case %d: ",z)
#define LINE_PRINT_CASE printf("Case %d:\n",z)
#define CASE_PRINT cout<<"Case "<<z<<": "
#define all(a) a.begin(),a.end()
#define intlim 2147483648
#define infinity (1<<28)
#define ull unsigned long long
#define gcd(a, b) __gcd(a, b)
#define lcm(a, b) ((a)*((b)/gcd(a,b)))
using namespace std;
//using namespace __gnu_pbds;
//typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1}; // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1}; // Kings Move
//const int fx[]={-2, -2, -1, -1, 1, 1, 2, 2}; // Knights Move
//const int fy[]={-1, 1, -2, 2, -2, 2, -1, 1}; // Knights Move
/*------------------------------------------------*/
/*-----------------------Bitmask------------------*/
//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/
#define mx 100005
#define segment_tree int l=n*2,r=l+1,mid=(b+e)/2
struct data
{
ll sum,sqrsum,lazy,upd;
};
data tree[3*mx];
ll ara[mx];
void push_down(int n, int b, int e)
{
segment_tree;
if(tree[n].upd)
{
tree[l].lazy=0;
tree[r].lazy=0;
tree[l].sum=(mid-b+1)*tree[n].upd;
tree[l].sqrsum=(mid-b+1)*sqr(tree[n].upd);
tree[r].sum=(e-mid)*tree[n].upd;
tree[r].sqrsum=(e-mid)*sqr(tree[n].upd);
tree[l].upd=tree[n].upd;
tree[r].upd=tree[n].upd;
tree[n].upd=0;
}
if(tree[n].lazy)
{
// tree[l].lazy+=tree[n].lazy;
// tree[l].status=1;
// tree[r].lazy+=tree[n].lazy;
// tree[r].status=1;
// tree[n].status=0;
tree[l].sqrsum+=(tree[l].sum*(2*tree[n].lazy))+(mid-b+1)*sqr(tree[n].lazy);
tree[r].sqrsum+=(tree[r].sum*(2*tree[n].lazy))+(e-mid)*sqr(tree[n].lazy);
tree[l].sum+=(mid-b+1)*tree[n].lazy;
tree[r].sum+=(e-mid)*tree[n].lazy;
tree[l].lazy+=tree[n].lazy;
tree[r].lazy+=tree[n].lazy;
tree[n].lazy=0;
}
}
void init(int n, int b, int e)
{
if(b==e)
{
tree[n].sum=ara[b];
tree[n].sqrsum=sqr(ara[b]);
tree[n].lazy=0;
tree[n].upd=0;
return;
}
segment_tree;
init(l,b,mid);
init(r,mid+1,e);
tree[n].sum=tree[l].sum+tree[r].sum;
tree[n].sqrsum=tree[l].sqrsum+tree[r].sqrsum;
tree[n].lazy=0;
tree[n].upd=0;
}
void update(int n, int b, int e, int i, int j, ll val, int type)
{
if(b>j || e<i) return;
if(b>=i && e<=j)
{
if(type==0)
{
tree[n].sum=(e-b+1)*val;
tree[n].sqrsum=(e-b+1)*sqr(val);
tree[n].lazy=0;
tree[n].upd=val;
}
else
{
tree[n].sqrsum+=(tree[n].sum*2*val)+((e-b+1)*sqr(val));
tree[n].sum+=(e-b+1)*val;
tree[n].lazy+=val;
}
return;
}
push_down(n,b,e);
segment_tree;
update(l,b,mid,i,j,val,type);
update(r,mid+1,e,i,j,val,type);
tree[n].sum=tree[l].sum+tree[r].sum;
tree[n].sqrsum=tree[l].sqrsum+tree[r].sqrsum;
}
ll query(int n, int b, int e, int i, int j)
{
if(b>j || e<i) return 0;
if(b>=i && e<=j)
return tree[n].sqrsum;
push_down(n,b,e);
segment_tree;
ll p=query(l,b,mid,i,j);
ll q=query(r,mid+1,e,i,j);
return p+q;
}
int main()
{
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
int t;
sf(t);
TEST_CASE(t)
{
int n,q;
sff(n,q);
loop1(i,n) sfl(ara[i]);
init(1,1,n);
LINE_PRINT_CASE;
while(q--)
{
int a,b,c,d;
sfff(a,b,c);
if(a==2)
{
ll ans=query(1,1,n,b,c);
printf("%lld\n",ans);
}
else
{
sf(d);
if(a==1)
update(1,1,n,b,c,d,1);
else
update(1,1,n,b,c,d,0);
}
}
}
return 0;
}