EXPWEI - EDITORIAL

PROBLEM LINK:

Practice
Contest: Division 1
Contest: Division 2
Contest: Division 3

Author: Jyothi Surya Prakash Bugatha
Tester: Aryan Choudhary
Editorialist: Nandini Kapoor

DIFFICULTY:

Cakewalk

PREREQUISITES:

Math, Fermats’s Little Theorem, Combinatorics

PROBLEM:

Given a sequence A of length N, the weight of this sequence is defined as weight(A)=\sum_{i=1}^{N}i⋅A_i. Given an integer N, we pick a permutation A of [1,2…N] uniformly at random. What is the expected weight of this permutation? It can be shown that this can be represented as P/Q where P and Q are co prime integers and Q≠0 mod 998244353. Print the value of P⋅Q−1 modulo 998244353.

EXPLANATION:

weight(A) for a permutation A from 1 to N as mentioned in the problem is given by -

weight(A) = 1\times A_1 + 2\times A_2 + \dots + n\times A_n

This in turn is calculated for all possible permutations A, let the evaluation of the value weight(A) for the i_{th} permutation be w_i (where 0\lt i\leq N!). We are required to calculate the expected value of weight of such a randomly picked permutation, this will be given by -

E = (w_1+w_2+\dots +w_{n!})/(n!)

If we fix the i_{th} element of A to a value j (where 0\lt j\leq N), there will be (N-1)! permutations possible by changing positions of other elements. In simpler words, there will be (N-1)! weights in E whose corresponding permutation A had the number j at the i_{th} position, meaning i\times j is contributing its value (N-1)! times to E.

For example consider N=3, in this case if we fix A_i=j for j=1 and i from 1 to N we observe -

  1. i=1, j=1
    A_1=1 -
    A=[1, 2, 3], weight(A)=1+2\times 2+3\times 3
    A=[1, 3, 2], weight(A)=1+2\times 3+3\times 2

  2. i=2, j=1
    A_2=1 -
    A=[2, 1, 3], weight(A)=2+2\times 1+3\times 3
    A=[3, 1, 2], weight(A)=3+2\times 1+3\times 2

  3. i=3, j=1
    A_3=1 -
    A=[2, 3, 1], weight(A)=2+2\times 3+3\times 1
    A=[3, 2, 1], weight(A)=3+2\times 2+3\times 1

We can observe that for each value of i from 1 to N, i\times j= i\times 1 is present in the weight of (N-1)!=(3-1)!=2 permutations of A.

Now we’ve established that each number j will appear at a position i (0<i\leq N), (N-1)! times. Each number j from 1 to N can contribute i\times j to the expected value (N-1)! times. We can write it as follows -

1\times i\times (N-1)! + 2\times i\times (N-1)! + 3\times i\times (N-1)! + \dots N\times i\times (N-1)!
=i\times (N-1)!\times (1+2+3+\dots +N)
=i\times (N-1)!\times ((N)(N+1)/2)

where i can assume values between 1 to N for each value of j as well, we can write this as -

1\times (N-1)!\times ((N)(N+1)/2) + 2\times (N-1)!\times ((N)(N+1)/2) + 3\times (N-1)!\times ((N)(N+1)/2) + \dots N\times (N-1)!\times ((N)(N+1)/2)
= (1+2+3+\dots +N)\times (N-1)!\times ((N)(N+1)/2)
= (N\times (N+1)/2)\times (N-1)!\times ((N)(N+1)/2)
= (N!\times N\times (N+1)^2)/4

We have above obtained the sum of weights of all N! permutations of A, to calculate the expected value E now we need to divide this sum by the number of permutations possible -

(N!\times N\times (N+1)^2)/(4\times N!)
=(N\times (N+1)^2)/4

Out of the components of the answer above, N\times (N+1)/2 will always yield an integral quotient which is why we won’t be taking the modular multiplicative inverse of 4 that is currently in the denominator. We can divide it into 2 parts -

  1. The part yielding integral quotient
  2. The part that may or may not leave a remainder on division

as follows -

(N\times (N+1)/2)
\times
(N+1)/2

If we compare it to P/Q form where P and Q are co prime integers, we can claim that either

  1. P=N\times (N+1)^2/2 and Q=2 or
  2. P=N\times (N+1)^2/4 and Q=1

Now if N+1 is found to be divisible by 2, we can simply multiply P by 1^{−1} modulo (998244353). On the other hand if P is not divisible by 2, we will have to multiply it with 2^{−1} modulo (998244353). This leads us to our final answer -

(N(N+1)^2/2)\times 2^{-1}modulo(998244353) if N is even, or
(N(N+1)^2/4)\times 1^{-1}modulo(998244353) if N is odd

TIME COMPLEXITY:

O(1) per test case, because computing modular multiplicative inverse i.e. Q^{-1}modulo(998244353) takes O(log(998244353)) time which is a constant.

SOLUTIONS:

Setter
Tester
/* in the name of Anton */

/*
  Compete against Yourself.
  Author - Aryan (@aryanc403)
  Atcoder library - https://atcoder.github.io/ac-library/production/document_en/
*/

#ifdef ARYANC403
    #include <header.h>
#else
    #pragma GCC optimize ("Ofast")
    #pragma GCC target ("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
    //#pragma GCC optimize ("-ffloat-store")
    #include<bits/stdc++.h>
    #define dbg(args...) 42;
#endif

// y_combinator from @neal template https://codeforces.com/contest/1553/submission/123849801
// http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2016/p0200r0.html
template<class Fun> class y_combinator_result {
    Fun fun_;
public:
    template<class T> explicit y_combinator_result(T &&fun): fun_(std::forward<T>(fun)) {}
    template<class ...Args> decltype(auto) operator()(Args &&...args) { return fun_(std::ref(*this), std::forward<Args>(args)...); }
};
template<class Fun> decltype(auto) y_combinator(Fun &&fun) { return y_combinator_result<std::decay_t<Fun>>(std::forward<Fun>(fun)); }

using namespace std;
#define fo(i,n)   for(i=0;i<(n);++i)
#define repA(i,j,n)   for(i=(j);i<=(n);++i)
#define repD(i,j,n)   for(i=(j);i>=(n);--i)
#define all(x) begin(x), end(x)
#define sz(x) ((lli)(x).size())
#define pb push_back
#define mp make_pair
#define X first
#define Y second
#define endl "\n"

typedef long long int lli;
typedef long double mytype;
typedef pair<lli,lli> ii;
typedef vector<ii> vii;
typedef vector<lli> vi;

const auto start_time = std::chrono::high_resolution_clock::now();
void aryanc403()
{
#ifdef ARYANC403
auto end_time = std::chrono::high_resolution_clock::now();
std::chrono::duration<double> diff = end_time-start_time;
    cerr<<"Time Taken : "<<diff.count()<<"\n";
#endif
}

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) {
    return readInt(l,r,' ');
}
long long readIntLn(long long l, long long r) {
    return readInt(l,r,'\n');
}
string readStringLn(int l, int r) {
    return readString(l,r,'\n');
}
string readStringSp(int l, int r) {
    return readString(l,r,' ');
}

void readEOF(){
    assert(getchar()==EOF);
}

vi readVectorInt(int n,lli l,lli r){
    vi a(n);
    for(int i=0;i<n-1;++i)
        a[i]=readIntSp(l,r);
    a[n-1]=readIntLn(l,r);
    return a;
}

// #include<atcoder/dsu>
// vector<vi> readTree(const int n){
//     vector<vi> e(n);
//     atcoder::dsu d(n);
//     for(lli i=1;i<n;++i){
//         const lli u=readIntSp(1,n)-1;
//         const lli v=readIntLn(1,n)-1;
//         e[u].pb(v);
//         e[v].pb(u);
//         d.merge(u,v);
//     }
//     assert(d.size(0)==n);
//     return e;
// }

const lli INF = 0xFFFFFFFFFFFFFFFL;

lli seed;
mt19937 rng(seed=chrono::steady_clock::now().time_since_epoch().count());
inline lli rnd(lli l=0,lli r=INF)
{return uniform_int_distribution<lli>(l,r)(rng);}

class CMP
{public:
bool operator()(ii a , ii b) //For min priority_queue .
{    return ! ( a.X < b.X || ( a.X==b.X && a.Y <= b.Y ));   }};

void add( map<lli,lli> &m, lli x,lli cnt=1)
{
    auto jt=m.find(x);
    if(jt==m.end())         m.insert({x,cnt});
    else                    jt->Y+=cnt;
}

void del( map<lli,lli> &m, lli x,lli cnt=1)
{
    auto jt=m.find(x);
    if(jt->Y<=cnt)            m.erase(jt);
    else                      jt->Y-=cnt;
}

bool cmp(const ii &a,const ii &b)
{
    return a.X<b.X||(a.X==b.X&&a.Y<b.Y);
}

const lli mod = 1000000007L;
// const lli maxN = 1000000007L;


#include <cassert>
#include <numeric>
#include <type_traits>

#ifdef _MSC_VER
#include <intrin.h>
#endif


#include <utility>

#ifdef _MSC_VER
#include <intrin.h>
#endif

namespace atcoder {

namespace internal {

constexpr long long safe_mod(long long x, long long m) {
    x %= m;
    if (x < 0) x += m;
    return x;
}

struct barrett {
    unsigned int _m;
    unsigned long long im;

    explicit barrett(unsigned int m) : _m(m), im((unsigned long long)(-1) / m + 1) {}

    unsigned int umod() const { return _m; }

    unsigned int mul(unsigned int a, unsigned int b) const {

        unsigned long long z = a;
        z *= b;
#ifdef _MSC_VER
        unsigned long long x;
        _umul128(z, im, &x);
#else
        unsigned long long x =
            (unsigned long long)(((unsigned __int128)(z)*im) >> 64);
#endif
        unsigned int v = (unsigned int)(z - x * _m);
        if (_m <= v) v += _m;
        return v;
    }
};

constexpr long long pow_mod_constexpr(long long x, long long n, int m) {
    if (m == 1) return 0;
    unsigned int _m = (unsigned int)(m);
    unsigned long long r = 1;
    unsigned long long y = safe_mod(x, m);
    while (n) {
        if (n & 1) r = (r * y) % _m;
        y = (y * y) % _m;
        n >>= 1;
    }
    return r;
}

constexpr bool is_prime_constexpr(int n) {
    if (n <= 1) return false;
    if (n == 2 || n == 7 || n == 61) return true;
    if (n % 2 == 0) return false;
    long long d = n - 1;
    while (d % 2 == 0) d /= 2;
    constexpr long long bases[3] = {2, 7, 61};
    for (long long a : bases) {
        long long t = d;
        long long y = pow_mod_constexpr(a, t, n);
        while (t != n - 1 && y != 1 && y != n - 1) {
            y = y * y % n;
            t <<= 1;
        }
        if (y != n - 1 && t % 2 == 0) {
            return false;
        }
    }
    return true;
}
template <int n> constexpr bool is_prime = is_prime_constexpr(n);

constexpr std::pair<long long, long long> inv_gcd(long long a, long long b) {
    a = safe_mod(a, b);
    if (a == 0) return {b, 0};

    long long s = b, t = a;
    long long m0 = 0, m1 = 1;

    while (t) {
        long long u = s / t;
        s -= t * u;
        m0 -= m1 * u;  // |m1 * u| <= |m1| * s <= b


        auto tmp = s;
        s = t;
        t = tmp;
        tmp = m0;
        m0 = m1;
        m1 = tmp;
    }
    if (m0 < 0) m0 += b / s;
    return {s, m0};
}

constexpr int primitive_root_constexpr(int m) {
    if (m == 2) return 1;
    if (m == 167772161) return 3;
    if (m == 469762049) return 3;
    if (m == 754974721) return 11;
    if (m == 998244353) return 3;
    int divs[20] = {};
    divs[0] = 2;
    int cnt = 1;
    int x = (m - 1) / 2;
    while (x % 2 == 0) x /= 2;
    for (int i = 3; (long long)(i)*i <= x; i += 2) {
        if (x % i == 0) {
            divs[cnt++] = i;
            while (x % i == 0) {
                x /= i;
            }
        }
    }
    if (x > 1) {
        divs[cnt++] = x;
    }
    for (int g = 2;; g++) {
        bool ok = true;
        for (int i = 0; i < cnt; i++) {
            if (pow_mod_constexpr(g, (m - 1) / divs[i], m) == 1) {
                ok = false;
                break;
            }
        }
        if (ok) return g;
    }
}
template <int m> constexpr int primitive_root = primitive_root_constexpr(m);

unsigned long long floor_sum_unsigned(unsigned long long n,
                                      unsigned long long m,
                                      unsigned long long a,
                                      unsigned long long b) {
    unsigned long long ans = 0;
    while (true) {
        if (a >= m) {
            ans += n * (n - 1) / 2 * (a / m);
            a %= m;
        }
        if (b >= m) {
            ans += n * (b / m);
            b %= m;
        }

        unsigned long long y_max = a * n + b;
        if (y_max < m) break;
        n = (unsigned long long)(y_max / m);
        b = (unsigned long long)(y_max % m);
        std::swap(m, a);
    }
    return ans;
}

}  // namespace internal

}  // namespace atcoder


#include <cassert>
#include <numeric>
#include <type_traits>

namespace atcoder {

namespace internal {

#ifndef _MSC_VER
template <class T>
using is_signed_int128 =
    typename std::conditional<std::is_same<T, __int128_t>::value ||
                                  std::is_same<T, __int128>::value,
                              std::true_type,
                              std::false_type>::type;

template <class T>
using is_unsigned_int128 =
    typename std::conditional<std::is_same<T, __uint128_t>::value ||
                                  std::is_same<T, unsigned __int128>::value,
                              std::true_type,
                              std::false_type>::type;

template <class T>
using make_unsigned_int128 =
    typename std::conditional<std::is_same<T, __int128_t>::value,
                              __uint128_t,
                              unsigned __int128>;

template <class T>
using is_integral = typename std::conditional<std::is_integral<T>::value ||
                                                  is_signed_int128<T>::value ||
                                                  is_unsigned_int128<T>::value,
                                              std::true_type,
                                              std::false_type>::type;

template <class T>
using is_signed_int = typename std::conditional<(is_integral<T>::value &&
                                                 std::is_signed<T>::value) ||
                                                    is_signed_int128<T>::value,
                                                std::true_type,
                                                std::false_type>::type;

template <class T>
using is_unsigned_int =
    typename std::conditional<(is_integral<T>::value &&
                               std::is_unsigned<T>::value) ||
                                  is_unsigned_int128<T>::value,
                              std::true_type,
                              std::false_type>::type;

template <class T>
using to_unsigned = typename std::conditional<
    is_signed_int128<T>::value,
    make_unsigned_int128<T>,
    typename std::conditional<std::is_signed<T>::value,
                              std::make_unsigned<T>,
                              std::common_type<T>>::type>::type;

#else

template <class T> using is_integral = typename std::is_integral<T>;

template <class T>
using is_signed_int =
    typename std::conditional<is_integral<T>::value && std::is_signed<T>::value,
                              std::true_type,
                              std::false_type>::type;

template <class T>
using is_unsigned_int =
    typename std::conditional<is_integral<T>::value &&
                                  std::is_unsigned<T>::value,
                              std::true_type,
                              std::false_type>::type;

template <class T>
using to_unsigned = typename std::conditional<is_signed_int<T>::value,
                                              std::make_unsigned<T>,
                                              std::common_type<T>>::type;

#endif

template <class T>
using is_signed_int_t = std::enable_if_t<is_signed_int<T>::value>;

template <class T>
using is_unsigned_int_t = std::enable_if_t<is_unsigned_int<T>::value>;

template <class T> using to_unsigned_t = typename to_unsigned<T>::type;

}  // namespace internal

}  // namespace atcoder


namespace atcoder {

namespace internal {

struct modint_base {};
struct static_modint_base : modint_base {};

template <class T> using is_modint = std::is_base_of<modint_base, T>;
template <class T> using is_modint_t = std::enable_if_t<is_modint<T>::value>;

}  // namespace internal

template <int m, std::enable_if_t<(1 <= m)>* = nullptr>
struct static_modint : internal::static_modint_base {
    using mint = static_modint;

  public:
    static constexpr int mod() { return m; }
    static mint raw(int v) {
        mint x;
        x._v = v;
        return x;
    }

    static_modint() : _v(0) {}
    template <class T, internal::is_signed_int_t<T>* = nullptr>
    static_modint(T v) {
        long long x = (long long)(v % (long long)(umod()));
        if (x < 0) x += umod();
        _v = (unsigned int)(x);
    }
    template <class T, internal::is_unsigned_int_t<T>* = nullptr>
    static_modint(T v) {
        _v = (unsigned int)(v % umod());
    }

    unsigned int val() const { return _v; }

    mint& operator++() {
        _v++;
        if (_v == umod()) _v = 0;
        return *this;
    }
    mint& operator--() {
        if (_v == 0) _v = umod();
        _v--;
        return *this;
    }
    mint operator++(int) {
        mint result = *this;
        ++*this;
        return result;
    }
    mint operator--(int) {
        mint result = *this;
        --*this;
        return result;
    }

    mint& operator+=(const mint& rhs) {
        _v += rhs._v;
        if (_v >= umod()) _v -= umod();
        return *this;
    }
    mint& operator-=(const mint& rhs) {
        _v -= rhs._v;
        if (_v >= umod()) _v += umod();
        return *this;
    }
    mint& operator*=(const mint& rhs) {
        unsigned long long z = _v;
        z *= rhs._v;
        _v = (unsigned int)(z % umod());
        return *this;
    }
    mint& operator/=(const mint& rhs) { return *this = *this * rhs.inv(); }

    mint operator+() const { return *this; }
    mint operator-() const { return mint() - *this; }

    mint pow(long long n) const {
        assert(0 <= n);
        mint x = *this, r = 1;
        while (n) {
            if (n & 1) r *= x;
            x *= x;
            n >>= 1;
        }
        return r;
    }
    mint inv() const {
        if (prime) {
            assert(_v);
            return pow(umod() - 2);
        } else {
            auto eg = internal::inv_gcd(_v, m);
            assert(eg.first == 1);
            return eg.second;
        }
    }

    friend mint operator+(const mint& lhs, const mint& rhs) {
        return mint(lhs) += rhs;
    }
    friend mint operator-(const mint& lhs, const mint& rhs) {
        return mint(lhs) -= rhs;
    }
    friend mint operator*(const mint& lhs, const mint& rhs) {
        return mint(lhs) *= rhs;
    }
    friend mint operator/(const mint& lhs, const mint& rhs) {
        return mint(lhs) /= rhs;
    }
    friend bool operator==(const mint& lhs, const mint& rhs) {
        return lhs._v == rhs._v;
    }
    friend bool operator!=(const mint& lhs, const mint& rhs) {
        return lhs._v != rhs._v;
    }

  private:
    unsigned int _v;
    static constexpr unsigned int umod() { return m; }
    static constexpr bool prime = internal::is_prime<m>;
};

template <int id> struct dynamic_modint : internal::modint_base {
    using mint = dynamic_modint;

  public:
    static int mod() { return (int)(bt.umod()); }
    static void set_mod(int m) {
        assert(1 <= m);
        bt = internal::barrett(m);
    }
    static mint raw(int v) {
        mint x;
        x._v = v;
        return x;
    }

    dynamic_modint() : _v(0) {}
    template <class T, internal::is_signed_int_t<T>* = nullptr>
    dynamic_modint(T v) {
        long long x = (long long)(v % (long long)(mod()));
        if (x < 0) x += mod();
        _v = (unsigned int)(x);
    }
    template <class T, internal::is_unsigned_int_t<T>* = nullptr>
    dynamic_modint(T v) {
        _v = (unsigned int)(v % mod());
    }

    unsigned int val() const { return _v; }

    mint& operator++() {
        _v++;
        if (_v == umod()) _v = 0;
        return *this;
    }
    mint& operator--() {
        if (_v == 0) _v = umod();
        _v--;
        return *this;
    }
    mint operator++(int) {
        mint result = *this;
        ++*this;
        return result;
    }
    mint operator--(int) {
        mint result = *this;
        --*this;
        return result;
    }

    mint& operator+=(const mint& rhs) {
        _v += rhs._v;
        if (_v >= umod()) _v -= umod();
        return *this;
    }
    mint& operator-=(const mint& rhs) {
        _v += mod() - rhs._v;
        if (_v >= umod()) _v -= umod();
        return *this;
    }
    mint& operator*=(const mint& rhs) {
        _v = bt.mul(_v, rhs._v);
        return *this;
    }
    mint& operator/=(const mint& rhs) { return *this = *this * rhs.inv(); }

    mint operator+() const { return *this; }
    mint operator-() const { return mint() - *this; }

    mint pow(long long n) const {
        assert(0 <= n);
        mint x = *this, r = 1;
        while (n) {
            if (n & 1) r *= x;
            x *= x;
            n >>= 1;
        }
        return r;
    }
    mint inv() const {
        auto eg = internal::inv_gcd(_v, mod());
        assert(eg.first == 1);
        return eg.second;
    }

    friend mint operator+(const mint& lhs, const mint& rhs) {
        return mint(lhs) += rhs;
    }
    friend mint operator-(const mint& lhs, const mint& rhs) {
        return mint(lhs) -= rhs;
    }
    friend mint operator*(const mint& lhs, const mint& rhs) {
        return mint(lhs) *= rhs;
    }
    friend mint operator/(const mint& lhs, const mint& rhs) {
        return mint(lhs) /= rhs;
    }
    friend bool operator==(const mint& lhs, const mint& rhs) {
        return lhs._v == rhs._v;
    }
    friend bool operator!=(const mint& lhs, const mint& rhs) {
        return lhs._v != rhs._v;
    }

  private:
    unsigned int _v;
    static internal::barrett bt;
    static unsigned int umod() { return bt.umod(); }
};
template <int id> internal::barrett dynamic_modint<id>::bt(998244353);

using modint998244353 = static_modint<998244353>;
using modint1000000007 = static_modint<1000000007>;
using modint = dynamic_modint<-1>;

namespace internal {

template <class T>
using is_static_modint = std::is_base_of<internal::static_modint_base, T>;

template <class T>
using is_static_modint_t = std::enable_if_t<is_static_modint<T>::value>;

template <class> struct is_dynamic_modint : public std::false_type {};
template <int id>
struct is_dynamic_modint<dynamic_modint<id>> : public std::true_type {};

template <class T>
using is_dynamic_modint_t = std::enable_if_t<is_dynamic_modint<T>::value>;

}  // namespace internal

}  // namespace atcoder

using namespace atcoder;
using mint = modint998244353;
//using mint = modint1000000007;
using vm = vector<mint>;
std::ostream& operator << (std::ostream& out, const mint& rhs) {
        return out<<rhs.val();
    }

    lli T,n,i,j,k,in,cnt,l,r,u,v,x,y;
    lli m;
    string s;
    vi a;
    //priority_queue < ii , vector < ii > , CMP > pq;// min priority_queue .

int main(void) {
    ios_base::sync_with_stdio(false);cin.tie(NULL);
    // freopen("txt.in", "r", stdin);
    // freopen("txt.out", "w", stdout);
// cout<<std::fixed<<std::setprecision(35);
T=readIntLn(1,2e5);
while(T--)
{

    const int n=readIntLn(1,1e9);
    mint ans=1;
    ans*=n;
    ans*=n+1;
    ans*=n+1;
    ans/=4;
    cout<<ans<<endl;
}   aryanc403();
    readEOF();
    return 0;
}
Editorialist
#include<bits/stdc++.h>
using namespace std;

#define _z ios_base::sync_with_stdio(false); cin.tie(NULL);
#define endl "\n"
#define mod 998244353
#define int long long

//______________________________z_____________________________

int power(int x, int y, int m)
{
    if(y == 0) return 1;
    int p = power(x, y / 2, m) % m;
    p = (p * p) % m;
    if(y % 2 == 0) return p;
    else return (x * p) % m;
}

int modInverse(int a, int m)
{
    int g = __gcd(a, m);
    return power(a, m - 2, m);
}

void solve() {
    int n, k;
    cin >> n;
    int num = n * (n + 1) / 2;
    num %= mod;
    if((n + 1) % 2 != 0) num *= (n + 1);
    else num *= (n + 1) / 2;
    num %= mod;
    int modinv = modInverse((n + 1) % 2 + 1, mod);
    cout << (modinv * num) % mod << endl;
}

int32_t main()
{
    _z;
    int t=1;
    cin>>t;
    while(t--) {
        solve();
    }
}

6 Likes

Beautiful Explanation :slightly_smiling_face:

2 Likes

why is this WA, i did the same thing.
https://www.codechef.com/viewsolution/54417400

2 Likes

The maximum value of long in java is 2^{63} which is 10^{19} for n=10^8 you numerator won’t have correct values. Print its value and see. This type of bug is often called int overflow.

To get an AC replace that num calculation line into 3 lines taking mod each time. Like -

long num=1; 
num=(num*n)%mod;
num=(num*n)%mod;
num=(num*(n+1))%mod;
4 Likes

Nice Explanation buddy !

1 Like

This problem could be solved in a less tedious manner by using Linearity of Expectation.

my c++ solution still gives wrong answer even after handling these overflows
solution:
my code

You have got an AC so I’m assuming you were able to fix the issue.

ya i guess i got it … sorry for the bother

I reckon it will yield a long expression with cancellations or summation formulas to be applied as well, I may be wrong though since I didn’t think of it from that perspective yet. Anywhooo I think it’s always wonderful when people propose different solutions, almost always insightful :blush:

1 Like

Can anyone please tell me why I am getting wrong answer.
My Solution link:- Solution: 54566593 | CodeChef

Hi @aryanc403 I modified my submission as you suggested but still getting WA on Task 2.
Can you let me know why.

Code : Solution: 54628143 | CodeChef

1 Like

Lol I did not even come close to the Author’s solution. I solved it during practice and the first 10 mins I was just thinking how is this the First problem ??!!

So I made some observations and found a pattern in the answer.

For N = 1, 2, 3, 4, 5, 6, 7…
The Answer was 1, 4.5, 12, 25, 45, 73.5, 112…

Then I found that this is coming out to be a Cubic Progression.

1---------- 4.5---------- 12---------- 25 ---------- 45 ---------- 73.5 ---------- 112
----3.5 ---------- 7.5 ---------- 13 ---------- 20---------- 28.5 ---------- 38.5
-------------- 4 ---------- 5.5---------- 7---------- 8.5------------- 10
-------------------1.5---------- 1.5 ---------- 1.5 ---------- 1.5 <----- Final Difference

And with that I found how to solve it with the image below.

My Solution :- Solution

3 Likes