Divisible Pairs CodeChef Solution

Problem -Divisible Pairs CodeChef Solution

This website is dedicated for CodeChef solution where we will publish right solution of all your favourite CodeChef problems along with detailed explanatory of different competitive programming concepts and languages.

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);}
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 readvall5(vall5& A) {rep(i,0,sz(A)) cin>>A[i][0]>>A[i][1]>>A[i][2]>>A[i][3]>>A[i][4];}
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;
    }

    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;
    T radius;
    Circle(T y, T x, T radius) : center(Point<T>(y,x)), radius(radius) {}
    Circle(Point<T> center, T radius) : center(center), 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) {
    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[]){
		InputReader in = new InputReader(System.in);
		OutputWriter out = new OutputWriter(System.out);
		int cases = in.readInt();
		for(int i=0;i<cases;i++){
	int N = in.readInt();
int M = in.readInt();
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();
	}
}
class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;

public InputReader(InputStream stream) {
this.stream = stream;
}

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

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

public String readString() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuffer res = new StringBuffer();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}

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

public String next() {
return readString();
}
}

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():
	for line in xrange(int(sys.stdin.readline().strip())):
		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()
Divisible Pairs CodeChef Solution Review:

In our experience, we suggest you solve this Divisible Pairs CodeChef Solution and gain some new skills from Professionals completely free and we assure you will be worth it.

If you are stuck anywhere between any coding problem, just visit Queslers to get the Divisible Pairs CodeChef Solution.

Find on CodeChef

Conclusion:

I hope this Divisible Pairs CodeChef Solution would be useful for you to learn something new from this problem. If it helped you then don’t forget to bookmark our site for more Coding Solutions.

This Problem is intended for audiences of all experiences who are interested in learning about Programming Language in a business context; there are no prerequisites.

Keep Learning!

More Coding Solutions >>

Cognitive Class Answer

CodeChef Solution

Microsoft Learn

Leave a Reply

Your email address will not be published. Required fields are marked *