Algorithm
Problem link- https://www.spoj.com/problems/HORRIBLE/
HORRIBLE - Horrible Queries
World is getting more evil and it's getting tougher to get into the Evil League of Evil. Since the legendary Bad Horse has retired, now you have to correctly answer the evil questions of Dr. Horrible, who has a PhD in horribleness (but not in Computer Science). You are given an array of N elements, which are initially all 0. After that you will be given C commands. They are -
* 0 p q v - you have to add v to all numbers in the range of p to q (inclusive), where p and q are two indexes of the array.
* 1 p q - output a line containing a single integer which is the sum of all the array elements between p and q (inclusive)
Input
In the first line you'll be given T, number of test cases.
Each test case will start with N (N <= 100 000) and C (C <= 100 000). After that you'll be given C commands in the format as mentioned above. 1 <= p, q <= N and 1 <= v <= 10^7.
Output
Print the answers of the queries.
Example
Input:
1
8 6
0 2 4 26
0 4 8 80
0 4 5 20
1 8 8
0 5 7 14
1 4 8
Output:
80
508
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MOD 1000000007LL
#define EPS 1e-9
#define io ios_base::sync_with_stdio(false);cin.tie(NULL);
#define M_PI 3.14159265358979323846
template <typename T> T gcd(T a, T b){return (b==0)?a:gcd(b,a%b);}
template <typename T> T lcm(T a, T b){return a*(b/gcd(a,b));}
template <typename T> T mod_exp(T b, T p, T m){T x = 1;while(p){if(p&1)x=(x*b)%m;b=(b*b)%m;p=p>>1;}return x;}
template <typename T> T invFermat(T a, T p){return mod_exp(a, p-2, p);}
template <typename T> T exp(T b, T p){T x = 1;while(p){if(p&1)x=(x*b);b=(b*b);p=p>>1;}return x;}
const int MAXN = 1e5+5;
ll BIT1[MAXN], BIT2[MAXN];
void update(int idx, ll val, int bit){
while(idx <= 100000){
if(bit == 1)
BIT1[idx] += val;
else
BIT2[idx] += val;
idx += idx&-idx;
}
}
void updateQuery(int l, int r, ll val){
update(l, val, 1);
update(r+1, -val, 1);
update(l, val*(l-1), 2);
update(r+1, -val*r, 2);
}
ll query(int idx, int bit){
ll sum = 0;
while(idx > 0){
sum += ((bit == 1) ? BIT1[idx] : BIT2[idx]);
idx -= idx&-idx;
}
return sum;
}
ll queryUtil(int idx){
return idx*query(idx, 1) - query(idx, 2);
}
ll queryRange(int l, int r){
return queryUtil(r) - queryUtil(l-1);
}
int main(){
io;
int t;
cin >> t;
while(t--){
memset(BIT1, 0, sizeof BIT1);
memset(BIT2, 0, sizeof BIT2);
int n, c;
cin >> n >> c;
for(int i = 0;i < c; i++){
int type;
cin >> type;
if(type == 0){
ll p, q, v;
cin >> p >> q >> v;
updateQuery(p, q, v);
}else{
int p, q;
cin >> p >> q;
cout << queryRange(p, q) << endl;
}
}
}
return 0;
}Input
8 6
0 2 4 26
0 4 8 80
0 4 5 20
1 8 8
0 5 7 14
1 4 8
Output
508
Demonstration
SPOJ Solution-Horrible Queries-Solution in C, C++, Java, Python