Substring Minimum Function

• Time : $O(1)$
• Space : $O(1)$

#include <bits/stdc++.h>

#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#pragma GCC optimize("Ofast")//Comment optimisations for interactive problems (use endl)
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")

using namespace std;

struct PairHash {inline std::size_t operator()(const std::pair<int, int> &v) const { return v.first * 31 + v.second; }};

// speed
#define Code ios_base::sync_with_stdio(false);
#define By ios::sync_with_stdio(0);
#define Sumfi cout.tie(NULL);

// alias
using ll = long long;
using ld = long double;
using ull = unsigned long long;

// constants
const ld PI = 3.14159265358979323846;  /* pi */
const ll INF = 1e18;
const ld EPS = 1e-4;
const ll MAX_N = 202020;
const ll mod = 998244353;

// typedef
typedef pair<ll, ll> pll;
typedef vector<pll> vpll;
typedef array<ll,3> all3;
typedef array<ll,5> all5;
typedef vector<all3> vall3;
typedef vector<all5> vall5;
typedef vector<ld> vld;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<int> vi;
typedef deque<ll> dll;
typedef deque<pll> dpll;
typedef pair<string, string> pss;
typedef vector<pss> vpss;
typedef vector<string> vs;
typedef vector<vs> vvs;
typedef unordered_set<ll> usll;
typedef unordered_set<pll, PairHash> uspll;
typedef unordered_map<ll, ll> umll;
typedef unordered_map<pll, ll, PairHash> umpll;

// macros
#define rep(i,m,n) for(ll i=m;i<n;i++)
#define rrep(i,m,n) for(ll i=n;i>=m;i--)
#define all(a) begin(a), end(a)
#define rall(a) rbegin(a), rend(a)
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
#define INF(a) memset(a,0x3f3f3f3f3f3f3f3fLL,sizeof(a))
#define ASCEND(a) iota(all(a),0)
#define sz(x) ll((x).size())
#define BIT(a,i) (a & (1ll<<i))
#define pyes cout<<"Yes\n";
#define pno cout<<"No\n";
#define pneg1 cout<<"-1\n";
#define ppossible cout<<"Possible\n";
#define pimpossible cout<<"Impossible\n";
#define TC(x) cout<<"Case #"<<x<<": ";
#define X first
#define Y second

// utility functions
template <typename T>
void print(T &&t)  { cout << t << "\n"; }
template<typename T>
void printv(vector<T>v){ll n=v.size();rep(i,0,n)cout<<v[i]<<" ";cout<<"\n";}
void fileIO() {freopen("input.txt","r",stdin); freopen("output.txt","w",stdout);}
void hackerCupIO() {freopen("/Users/summerflower/Downloads/worklife_balance_chapter_1_input (1).txt", "r", stdin); freopen("/Users/summerflower/Downloads/solution.txt","w",stdout);}
void readf() {freopen("", "rt", stdin);}
template<typename T>
void readv(vector<T>& v){rep(i,0,sz(v)) cin>>v[i];}
template<typename T, typename U>
void readp(pair<T,U>& A) {cin>>A.first>>A.second;}
template<typename T, typename U>
void readvp(vector<pair<T,U>>& A) {rep(i,0,sz(A)) readp(A[i]); }
void readvall3(vall3& A) {rep(i,0,sz(A)) cin>>A[i][0]>>A[i][1]>>A[i][2];}
void readvvll(vvll& A) {rep(i,0,sz(A)) readv(A[i]);}

struct Combination {
    vll fac, inv;
    ll n, MOD;

    ll modpow(ll n, ll x, ll MOD = mod) { if(!x) return 1; ll res = modpow(n,x>>1,MOD); res = (res * res) % MOD; if(x&1) res = (res * n) % MOD; return res; }

    Combination(ll _n, ll MOD = mod): n(_n + 1), MOD(MOD) {
        inv = fac = vll(n,1);
        rep(i,1,n) fac[i] = fac[i-1] * i % MOD;
        inv[n - 1] = modpow(fac[n - 1], MOD - 2, MOD);
        rrep(i,1,n - 2) inv[i] = inv[i + 1] * (i + 1) % MOD;
    }

    ll fact(ll n) {return fac[n];}
    ll nCr(ll n, ll r) {
        if(n < r or n < 0 or r < 0) return 0;
        return fac[n] * inv[r] % MOD * inv[n-r] % MOD;
    }
};

struct Matrix {
    ll r,c;
    vvll matrix;
    Matrix(ll r, ll c, ll v = 0): r(r), c(c), matrix(vvll(r,vll(c,v))) {}

    Matrix operator*(const Matrix& B) const {
        Matrix res(r, B.c);
        rep(i,0,r) rep(j,0,B.c) rep(k,0,B.r) {
                    res.matrix[i][j] = (res.matrix[i][j] + matrix[i][k] * B.matrix[k][j] % mod) % mod;
                }
        return res;
    }
};

// geometry data structures
template <typename T>
struct Point {
    T y,x;
    Point(T y, T x) : y(y), x(x) {}
    Point() {}
    void input() {cin>>y>>x;}
    friend ostream& operator<<(ostream& os, const Point<T>& p) { os<<p.y<<' '<<p.x<<'\n'; return os;}
    Point<T> operator+(Point<T>& p) {return Point<T>(y + p.y, x + p.x);}
    Point<T> operator-(Point<T>& p) {return Point<T>(y - p.y, x - p.x);}
    Point<T> operator*(ll n) {return Point<T>(y*n,x*n); }
    Point<T> operator/(ll n) {return Point<T>(y/n,x/n); }

    Point<T> rotate(Point<T> center, ld angle) {
        ld si = sin(angle * PI / 180.), co = cos(angle * PI / 180.);
        ld y = this->y - center.y;
        ld x = this->x - center.x;

        return Point<T>(y * co - x * si + center.y, y * si + x * co + center.x);
    }
};

template<typename T>
struct Circle {
    Point<T> center;
    T radius;
    Circle(T y, T x, T radius) : center(Point<T>(y,x)), radius(radius) {}
    Circle() {}

    void input() {
        center = Point<T>();
        center.input();
        cin>>radius;
    }

    bool circumference(Point<T> p) {
        return (center.x - p.x) * (center.x - p.x) + (center.y - p.y) * (center.y - p.y) == radius * radius;
    }

    bool intersect(Circle<T> c) {
        T d = (center.x - c.center.x) * (center.x - c.center.x) + (center.y - c.center.y) * (center.y - c.center.y);
        return (radius - c.radius) * (radius - c.radius) <= d and d <= (radius + c.radius) * (radius + c.radius);
    }

    bool include(Circle<T> c) {
        T d = (center.x - c.center.x) * (center.x - c.center.x) + (center.y - c.center.y) * (center.y - c.center.y);
        return d <= radius * radius;
    }
};

ll __gcd(ll x, ll y) { return !y ? x : __gcd(y, x % y); }
all3 __exgcd(ll x, ll y) { if(!y) return {x,1,0}; auto [g,x1,y1] = __exgcd(y, x % y); return {g, y1, x1 - (x/y) * y1}; }
ll __lcm(ll x, ll y) { return x / __gcd(x,y) * y; }
ll modpow(ll n, ll x, ll MOD = mod) { n%=MOD; if(!x) return 1; ll res = modpow(n,x>>1,MOD); res = (res * res) % MOD; if(x&1) res = (res * n) % MOD; return res; }

ll solve(ll n, ll m) {
    if(m * 2 + 1 >= n) return n - m;
    ll div = m + 1;
    ll quo = (n - m) / div, rem = (n - m) % div;
    ll res = (quo + 1) * (quo + 2) / 2 * rem + quo * (quo + 1) / 2 * (div - rem);
    return res;
}

int main() {
    Code By Sumfi
    cout.precision(12);
    ll tc = 1;
    cin>>tc;
    rep(i,1,tc+1) {
        ll n,m;
        cin>>n>>m;
        print(solve(n,m));
    }
    return 0;
}