# MANYSUMS - Editorial

Author: Hriday G
Tester: Rahul Dugar
Editorialist: Aman Dwivedi

Cakewalk

None

# PROBLEM:

Given a range [L,R] endpoints inclusive, You need to find the number of distinct pair sums such that the elements of pair belongs to the given range. Also the elements of the pair need not to be distinct.

# EXPLANATION:

We are given a range [L,R] endpoints inclusive. We need to find out the number of distinct integers possible such that it is the sum of any two not necessarily distinct integers from the given range.

The first observation that we can make is that the, number of distinct integers that are possible will lie in a continuous range since the given range is continuous.

Proof

Suppose we have a integer S that is the sum of two integers X and Y, where X and Y lies in a given range. Then, we can always obtain integers S-1 and S+1 by incrementing and decrementing either X or Y respectively.

Since X and Y, lies in a given range and the range is continuous, hence (X-1,X+1,Y-1,Y+1) will also lie in a given range. If we are unable to increment or decrement neither X nor Y, it means that they are endpoints of the given range and hence we say that S needs to be the endpoint of its range.

Now, as we know that the integers will be in continuous range hence we are only left with finding out the endpoints of the range. The endpoints of any continuous range are minimum and maximum integers possible.

• Here the Minimum integer that can be obtained will be 2*L.
• And the Maximum integer that can be obtained will be 2*R.

Since now as be know our range i.e. [2*L,2*R], we can easily count the number of integers in this range.

• Hence the number of distinct integers are (2*R-2*L+1).

O(1) per query

# SOLUTIONS:

Setter
``````/**
>> the_hyp0cr1t3
>> 05.01.2021 22:55:43
**/
#include <bits/stdc++.h>
using namespace std;
#define pb emplace_back
#define sz(x) int(x.size())
#define all(x) x.begin(), x.end()

const int64_t DESPACITO = 2e18;
const int INF = 2e9, MOD = 1e9+7;
const int N = 2e5 + 5;

int main() {
cin.tie(nullptr)->sync_with_stdio(false);
int Q; cin >> Q;
while(Q--) {
int l, r; cin >> l >> r;
cout << (r-l << 1) + 1 << '\n';
}
} // ~W
``````
Tester
``````#include <bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>
using namespace __gnu_pbds;
using namespace __gnu_cxx;
#ifndef rd
#define trace(...)
#define endl '\n'
#endif
#define pb push_back
#define fi first
#define se second
#define int long long
typedef long long ll;
typedef long double f80;
#define double long double
#define pii pair<int,int>
#define pll pair<ll,ll>
#define sz(x) ((long long)x.size())
#define fr(a,b,c) for(int a=b; a<=c; a++)
#define rep(a,b,c) for(int a=b; a<c; a++)
#define trav(a,x) for(auto &a:x)
#define all(con) con.begin(),con.end()
const ll infl=0x3f3f3f3f3f3f3f3fLL;
const int infi=0x3f3f3f3f;
//const int mod=998244353;
const int mod=1000000007;
typedef vector<int> vi;
typedef vector<ll> vl;

typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset;
auto clk=clock();
mt19937_64 rang(chrono::high_resolution_clock::now().time_since_epoch().count());
int rng(int lim) {
uniform_int_distribution<int> uid(0,lim-1);
return uid(rang);
}
int powm(int a, int b) {
int res=1;
while(b) {
if(b&1)
res=(res*a)%mod;
a=(a*a)%mod;
b>>=1;
}
return res;
}

long long readInt(long long l, long long r, char endd) {
long long x=0;
int cnt=0;
int fi=-1;
bool is_neg=false;
while(true) {
char g=getchar();
if(g=='-') {
assert(fi==-1);
is_neg=true;
continue;
}
if('0'<=g&&g<='9') {
x*=10;
x+=g-'0';
if(cnt==0) {
fi=g-'0';
}
cnt++;
assert(fi!=0 || cnt==1);
assert(fi!=0 || is_neg==false);

assert(!(cnt>19 || ( cnt==19 && fi>1) ));
} else if(g==endd) {
if(is_neg) {
x=-x;
}
assert(l<=x&&x<=r);
return x;
} else {
assert(false);
}
}
}
string readString(int l, int r, char endd) {
string ret="";
int cnt=0;
while(true) {
char g=getchar();
assert(g!=-1);
if(g==endd) {
break;
}
cnt++;
ret+=g;
}
assert(l<=cnt&&cnt<=r);
return ret;
}
long long readIntSp(long long l, long long r) {
}
long long readIntLn(long long l, long long r) {
}
string readStringLn(int l, int r) {
}
string readStringSp(int l, int r) {
}

void solve() {
cout<<(2*r-2*l+1)<<endl;
}

signed main() {
ios_base::sync_with_stdio(0),cin.tie(0);
srand(chrono::high_resolution_clock::now().time_since_epoch().count());
cout<<fixed<<setprecision(10);
//	int t;
//	cin>>t;
fr(i,1,t)
solve();
#ifdef rd
cerr<<endl<<endl<<endl<<"Time Elapsed: "<<((double)(clock()-clk))/CLOCKS_PER_SEC<<endl;
#endif
}

``````
Editorialist
``````#include<bits/stdc++.h>
using namespace std;

void solve(){
int l,r; cin>>l>>r;
l=2*l;
r=2*r;

int ans=r-l+1;

cout<<ans<<"\n";
}

int main(){
int t; cin>>t;
while(t--){
solve();
}

return 0;
}

``````