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# Divisible Pairs CodeChef Solution

## Divisible Pairs CodeChef Solution in C++14

``````#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-9;
const ll MAX_N = 1010101;
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> dqll;
typedef deque<pll> dqpll;
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 BITSHIFT(a,i,n) (((a<<i) & ((1ll<<n) - 1)) | (a>>(n-i)))
#define pyes cout<<"YES\n";
#define pno cout<<"NO\n";
#define endl "\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];if(i+1!=n)cout<<' ';}cout<<endl;}
template<typename T>
void printvln(vector<T>v){ll n=v.size();rep(i,0,n)cout<<v[i]<<endl;}
void fileIO(string in = "input.txt", string out = "output.txt") {freopen(in.c_str(),"r",stdin); freopen(out.c_str(),"w",stdout);}
template<typename T>
template<typename T, typename U>
template<typename T, typename U>

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;
}

Matrix copy() {
Matrix copy(r,c);
copy.matrix = matrix;
return copy;
}

Matrix pow(ll n) {
assert(r == c);
Matrix res(r,r);
Matrix now = copy();
rep(i,0,r) res.matrix[i][i] = 1;
while(n) {
if(n & 1) res = res * now;
now = now * now;
n /= 2;
}
return res;
}
};

// geometry data structures
template <typename T>
struct Point {
T y,x;
Point(T y, T x) : y(y), x(x) {}
Point(pair<T,T> p) : y(p.first), x(p.second) {}
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); }
bool operator<(const Point &other) const {if (x == other.x) return y < other.y;return x < other.x;}
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);
}
ld distance(Point<T> other) {
T dy = abs(this->y - other.y);
T dx = abs(this->x - other.x);
return sqrt(dy * dy + dx * dx);
}

T norm() { return x * x + y * y; }
};

template<typename T>
struct Line {
Point<T> A, B;
Line(Point<T> A, Point<T> B) : A(A), B(B) {}
Line() {}

void input() {
A = Point<T>();
B = Point<T>();
A.input();
B.input();
}

T ccw(Point<T> &a, Point<T> &b, Point<T> &c) {
T res = a.x * b.y + b.x * c.y + c.x * a.y;
res -= (a.x * c.y + b.x * a.y + c.x * b.y);
return res;
}

bool isIntersect(Line<T> o) {
T p1p2 = ccw(A,B,o.A) * ccw(A,B,o.B);
T p3p4 = ccw(o.A,o.B,A) * ccw(o.A,o.B,B);
if (p1p2 == 0 && p3p4 == 0) {
pair<T,T> p1(A.y, A.x), p2(B.y,B.x), p3(o.A.y, o.A.x), p4(o.B.y, o.B.x);
if (p1 > p2) swap(p2, p1);
if (p3 > p4) swap(p3, p4);
return p3 <= p2 && p1 <= p4;
}
return p1p2 <= 0 && p3p4 <= 0;
}

pair<bool,Point<ld>> intersection(Line<T> o) {
if(!this->intersection(o)) return {false, {}};
ld det = 1. * (o.B.y-o.A.y)*(B.x-A.x) - 1.*(o.B.x-o.A.x)*(B.y-A.y);
ld t = ((o.B.x-o.A.x)*(A.y-o.A.y) - (o.B.y-o.A.y)*(A.x-o.A.x)) / det;
return {true, {A.y + 1. * t * (B.y - A.y), B.x + 1. * t * (B.x - A.x)}};
}

//@formula for : y = ax + b
//@return {a,b};
pair<ld, ld> formula() {
T y1 = A.y, y2 = B.y;
T x1 = A.x, x2 = B.x;
if(y1 == y2) return {1e9, 0};
if(x1 == x2) return {0, 1e9};
ld a = 1. * (y2 - y1) / (x2 - x1);
ld b = -x1 * a + y1;
return {a, b};
}
};

template<typename T>
struct Circle {
Point<T> center;
Circle() {}

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

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);
}

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);
}
};

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) {
ll q = n / m, r = n % m;
if(n < m) return (r > m / 2) ? r - m / 2 : 0;
if(n == m) return (m - 1) / 2;
ll res = r * q + ((r > m / 2) ? r - m / 2 : 0);
if(m & 1) res += q * q * (m / 2) + q * (q - 1) / 2;
else res += q * q * ((m - 1) / 2) + q * (q - 1);
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;
}``````

## Divisible Pairs CodeChef Solution in PYTH 3

``````# cook your dish here

t = int(input())

for i in range(t):
n, m = map(int, input().split())
c = n // m
s = n % m
if m == 2:
print(((s + c) * (c + s - 1)) // 2 + (c * (c - 1)) // 2)
elif m % 2 == 0:
print((c * c * ((m - 1) // 2)) + ((c - 1) * c) + (c * s) + max(0, s - m // 2))
else:
print((c * c * ((m - 1) // 2)) + ((c - 1) * c) // 2 + (c * s) + max(0, s - m // 2))``````

## Divisible Pairs CodeChef Solution in C

``````
#include<stdio.h>
int main(){
long long t,n,m,i,d,c,j;
long long countAns;
long long x,y;

scanf("%lld",&t);
while(t--){
countAns=0;
scanf("%lld %lld",&n,&m);
x=n/m;
y=n%m;
countAns+=x*x*(m-1);
countAns+=x*(x-1);
if(m%2==0) countAns-=x;
countAns/=2;
countAns+=x*y;
if(y>m/2) countAns+=y-m/2;
printf("%lld\n",countAns);
}

return 0;
}``````

## Divisible Pairs CodeChef Solution in JAVA

``````import java.io.*;
import java.util.*;
import java.math.*;
class Main{
public static void main(String args[]){
OutputWriter out = new OutputWriter(System.out);
for(int i=0;i<cases;i++){
long x = N / M, y = N % M;
long ans = x * x * (M - 1);
ans += x * (x - 1);
if (M % 2 == 0) {
ans -= x;
}
ans /= 2;
ans += x * y;
if (y > M / 2)
ans += y - M / 2;
out.printLine(ans);
}
out.close();
}
}
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;

this.stream = stream;
}

if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}

while (isSpaceChar(c))
int sgn = 1;
if (c == '-') {
sgn = -1;
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
} while (!isSpaceChar(c));
return res * sgn;
}

while (isSpaceChar(c))
StringBuffer res = new StringBuffer();
do {
res.appendCodePoint(c);
} while (!isSpaceChar(c));
return res.toString();
}

static boolean isSpaceChar(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}

public String next() {
}
}

class OutputWriter {
private final PrintWriter writer;

public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(outputStream);
}

public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}

public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0)
writer.print(' ');
writer.print(objects[i]);
}
}

public void printLine(Object... objects) {
print(objects);
writer.println();
}

public void close() {
writer.close();
}
}``````

## Divisible Pairs CodeChef Solution in PYTH

``````import sys, psyco
psyco.full()
def main():
N, M = [int(i) for i in sys.stdin.readline().strip().split(' ')]
u, v = N/M, (2*N-1)/M
vu = v-u
if vu < 0:
vu = 0
if not M%2:
print (M*u*(u+1)-M*vu*(v+u+1))/4+(N-1)*vu-2*u+v
elif not u%2:
a = (M*u*(u+1)-7*u)/4
if vu>=1:
if not v%2:
l = vu/2
b = (l*(4*N-3-M*(2*u-1)))/2-M*l*(l+1)
else:
x, y = (vu-1)/2, (vu+1)/2
b = x*(N-1)+(y*(2*N-1)-M*(x*(x+u+1)+y*(y+u)))/2
print a+b+v
else:
print a+u
else:
m, n = (u-1)/2, (u+1)/2
a = M*(m*(m+1)+n*(n+1))/2-2*m-n*(M+3)/2
if vu >= 1:
if not v%2:
x, y = (vu-1)/2, (vu+1)/2
b = x*(N-1)+(y*(2*N-1)-M*(x*(x+u+1)+y*(y+u)))/2
else:
l = vu/2
b = (l*(4*N-3-M*(2*u-1)))/2-M*l*(l+1)
print a+b+v
else:
print a+u

main()``````
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