p193.ans1
======================
3

=================
p193.ans2
======================
1

=================
p193.cpp
======================
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

// base and base_digits must be consistent
const int base = 1000000000;
const int base_digits = 9;

struct bigint {
    vector<int> z;
    int sign;

    bigint() : sign(1) {}

    bigint(long long v) { *this = v; }

    bigint(const string& s) { read(s); }

    void operator=(const bigint& v) {
        sign = v.sign;
        z = v.z;
    }

    void operator=(long long v) {
        sign = 1;
        if(v < 0) {
            sign = -1, v = -v;
        }
        z.clear();
        for(; v > 0; v = v / base) {
            z.push_back(v % base);
        }
    }

    bigint operator+(const bigint& v) const {
        if(sign == v.sign) {
            bigint res = v;

            for(int i = 0, carry = 0;
                i < (int)max(z.size(), v.z.size()) || carry; ++i) {
                if(i == (int)res.z.size()) {
                    res.z.push_back(0);
                }
                res.z[i] += carry + (i < (int)z.size() ? z[i] : 0);
                carry = res.z[i] >= base;
                if(carry) {
                    res.z[i] -= base;
                }
            }
            return res;
        }
        return *this - (-v);
    }

    bigint operator-(const bigint& v) const {
        if(sign == v.sign) {
            if(abs() >= v.abs()) {
                bigint res = *this;
                for(int i = 0, carry = 0; i < (int)v.z.size() || carry; ++i) {
                    res.z[i] -= carry + (i < (int)v.z.size() ? v.z[i] : 0);
                    carry = res.z[i] < 0;
                    if(carry) {
                        res.z[i] += base;
                    }
                }
                res.trim();
                return res;
            }
            return -(v - *this);
        }
        return *this + (-v);
    }

    void operator*=(int v) {
        if(v < 0) {
            sign = -sign, v = -v;
        }
        for(int i = 0, carry = 0; i < (int)z.size() || carry; ++i) {
            if(i == (int)z.size()) {
                z.push_back(0);
            }
            long long cur = z[i] * (long long)v + carry;
            carry = (int)(cur / base);
            z[i] = (int)(cur % base);
            // asm("divl %%ecx" : "=a"(carry), "=d"(a[i]) : "A"(cur),
            // "c"(base));
        }
        trim();
    }

    bigint operator*(int v) const {
        bigint res = *this;
        res *= v;
        return res;
    }

    friend pair<bigint, bigint> divmod(const bigint& a1, const bigint& b1) {
        int norm = base / (b1.z.back() + 1);
        bigint a = a1.abs() * norm;
        bigint b = b1.abs() * norm;
        bigint q, r;
        q.z.resize(a.z.size());

        for(int i = a.z.size() - 1; i >= 0; i--) {
            r *= base;
            r += a.z[i];
            int s1 = b.z.size() < r.z.size() ? r.z[b.z.size()] : 0;
            int s2 = b.z.size() - 1 < r.z.size() ? r.z[b.z.size() - 1] : 0;
            int d = ((long long)s1 * base + s2) / b.z.back();
            r -= b * d;
            while(r < 0) {
                r += b, --d;
            }
            q.z[i] = d;
        }

        q.sign = a1.sign * b1.sign;
        r.sign = a1.sign;
        q.trim();
        r.trim();
        return make_pair(q, r / norm);
    }

    friend bigint sqrt(const bigint& a1) {
        bigint a = a1;
        while(a.z.empty() || a.z.size() % 2 == 1) {
            a.z.push_back(0);
        }

        int n = a.z.size();

        int firstDigit = (int)sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
        int norm = base / (firstDigit + 1);
        a *= norm;
        a *= norm;
        while(a.z.empty() || a.z.size() % 2 == 1) {
            a.z.push_back(0);
        }

        bigint r = (long long)a.z[n - 1] * base + a.z[n - 2];
        firstDigit = (int)sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
        int q = firstDigit;
        bigint res;

        for(int j = n / 2 - 1; j >= 0; j--) {
            for(;; --q) {
                bigint r1 =
                    (r - (res * 2 * base + q) * q) * base * base +
                    (j > 0 ? (long long)a.z[2 * j - 1] * base + a.z[2 * j - 2]
                           : 0);
                if(r1 >= 0) {
                    r = r1;
                    break;
                }
            }
            res *= base;
            res += q;

            if(j > 0) {
                int d1 =
                    res.z.size() + 2 < r.z.size() ? r.z[res.z.size() + 2] : 0;
                int d2 =
                    res.z.size() + 1 < r.z.size() ? r.z[res.z.size() + 1] : 0;
                int d3 = res.z.size() < r.z.size() ? r.z[res.z.size()] : 0;
                q = ((long long)d1 * base * base + (long long)d2 * base + d3) /
                    (firstDigit * 2);
            }
        }

        res.trim();
        return res / norm;
    }

    bigint operator/(const bigint& v) const { return divmod(*this, v).first; }

    bigint operator%(const bigint& v) const { return divmod(*this, v).second; }

    void operator/=(int v) {
        if(v < 0) {
            sign = -sign, v = -v;
        }
        for(int i = (int)z.size() - 1, rem = 0; i >= 0; --i) {
            long long cur = z[i] + rem * (long long)base;
            z[i] = (int)(cur / v);
            rem = (int)(cur % v);
        }
        trim();
    }

    bigint operator/(int v) const {
        bigint res = *this;
        res /= v;
        return res;
    }

    int operator%(int v) const {
        if(v < 0) {
            v = -v;
        }
        int m = 0;
        for(int i = z.size() - 1; i >= 0; --i) {
            m = (z[i] + m * (long long)base) % v;
        }
        return m * sign;
    }

    void operator+=(const bigint& v) { *this = *this + v; }
    void operator-=(const bigint& v) { *this = *this - v; }
    void operator*=(const bigint& v) { *this = *this * v; }
    void operator/=(const bigint& v) { *this = *this / v; }

    bool operator<(const bigint& v) const {
        if(sign != v.sign) {
            return sign < v.sign;
        }
        if(z.size() != v.z.size()) {
            return z.size() * sign < v.z.size() * v.sign;
        }
        for(int i = z.size() - 1; i >= 0; i--) {
            if(z[i] != v.z[i]) {
                return z[i] * sign < v.z[i] * sign;
            }
        }
        return false;
    }

    bool operator>(const bigint& v) const { return v < *this; }
    bool operator<=(const bigint& v) const { return !(v < *this); }
    bool operator>=(const bigint& v) const { return !(*this < v); }
    bool operator==(const bigint& v) const {
        return !(*this < v) && !(v < *this);
    }
    bool operator!=(const bigint& v) const { return *this < v || v < *this; }

    void trim() {
        while(!z.empty() && z.back() == 0) {
            z.pop_back();
        }
        if(z.empty()) {
            sign = 1;
        }
    }

    bool isZero() const { return z.empty() || (z.size() == 1 && !z[0]); }

    bigint operator-() const {
        bigint res = *this;
        res.sign = -sign;
        return res;
    }

    bigint abs() const {
        bigint res = *this;
        res.sign *= res.sign;
        return res;
    }

    long long longValue() const {
        long long res = 0;
        for(int i = z.size() - 1; i >= 0; i--) {
            res = res * base + z[i];
        }
        return res * sign;
    }

    friend bigint gcd(const bigint& a, const bigint& b) {
        return b.isZero() ? a : gcd(b, a % b);
    }
    friend bigint lcm(const bigint& a, const bigint& b) {
        return a / gcd(a, b) * b;
    }

    void read(const string& s) {
        sign = 1;
        z.clear();
        int pos = 0;
        while(pos < (int)s.size() && (s[pos] == '-' || s[pos] == '+')) {
            if(s[pos] == '-') {
                sign = -sign;
            }
            ++pos;
        }
        for(int i = s.size() - 1; i >= pos; i -= base_digits) {
            int x = 0;
            for(int j = max(pos, i - base_digits + 1); j <= i; j++) {
                x = x * 10 + s[j] - '0';
            }
            z.push_back(x);
        }
        trim();
    }

    friend istream& operator>>(istream& stream, bigint& v) {
        string s;
        stream >> s;
        v.read(s);
        return stream;
    }

    friend ostream& operator<<(ostream& stream, const bigint& v) {
        if(v.sign == -1) {
            stream << '-';
        }
        stream << (v.z.empty() ? 0 : v.z.back());
        for(int i = (int)v.z.size() - 2; i >= 0; --i) {
            stream << setw(base_digits) << setfill('0') << v.z[i];
        }
        return stream;
    }

    static vector<int> convert_base(
        const vector<int>& a, int old_digits, int new_digits
    ) {
        vector<long long> p(max(old_digits, new_digits) + 1);
        p[0] = 1;
        for(int i = 1; i < (int)p.size(); i++) {
            p[i] = p[i - 1] * 10;
        }
        vector<int> res;
        long long cur = 0;
        int cur_digits = 0;
        for(int i = 0; i < (int)a.size(); i++) {
            cur += a[i] * p[cur_digits];
            cur_digits += old_digits;
            while(cur_digits >= new_digits) {
                res.push_back(int(cur % p[new_digits]));
                cur /= p[new_digits];
                cur_digits -= new_digits;
            }
        }
        res.push_back((int)cur);
        while(!res.empty() && res.back() == 0) {
            res.pop_back();
        }
        return res;
    }

    typedef vector<long long> vll;

    static vll karatsubaMultiply(const vll& a, const vll& b) {
        int n = a.size();
        vll res(n + n);
        if(n <= 32) {
            for(int i = 0; i < n; i++) {
                for(int j = 0; j < n; j++) {
                    res[i + j] += a[i] * b[j];
                }
            }
            return res;
        }

        int k = n >> 1;
        vll a1(a.begin(), a.begin() + k);
        vll a2(a.begin() + k, a.end());
        vll b1(b.begin(), b.begin() + k);
        vll b2(b.begin() + k, b.end());

        vll a1b1 = karatsubaMultiply(a1, b1);
        vll a2b2 = karatsubaMultiply(a2, b2);

        for(int i = 0; i < k; i++) {
            a2[i] += a1[i];
        }
        for(int i = 0; i < k; i++) {
            b2[i] += b1[i];
        }

        vll r = karatsubaMultiply(a2, b2);
        for(int i = 0; i < (int)a1b1.size(); i++) {
            r[i] -= a1b1[i];
        }
        for(int i = 0; i < (int)a2b2.size(); i++) {
            r[i] -= a2b2[i];
        }

        for(int i = 0; i < (int)r.size(); i++) {
            res[i + k] += r[i];
        }
        for(int i = 0; i < (int)a1b1.size(); i++) {
            res[i] += a1b1[i];
        }
        for(int i = 0; i < (int)a2b2.size(); i++) {
            res[i + n] += a2b2[i];
        }
        return res;
    }

    bigint operator*(const bigint& v) const {
        vector<int> a6 = convert_base(this->z, base_digits, 6);
        vector<int> b6 = convert_base(v.z, base_digits, 6);
        vll a(a6.begin(), a6.end());
        vll b(b6.begin(), b6.end());
        while(a.size() < b.size()) {
            a.push_back(0);
        }
        while(b.size() < a.size()) {
            b.push_back(0);
        }
        while(a.size() & (a.size() - 1)) {
            a.push_back(0), b.push_back(0);
        }
        vll c = karatsubaMultiply(a, b);
        bigint res;
        res.sign = sign * v.sign;
        for(int i = 0, carry = 0; i < (int)c.size(); i++) {
            long long cur = c[i] + carry;
            res.z.push_back((int)(cur % 1000000));
            carry = (int)(cur / 1000000);
        }
        res.z = convert_base(res.z, 6, base_digits);
        res.trim();
        return res;
    }
};

bigint n;

void read() {
    cin >> n;
}

void solve() {
    // - The ball visits every girl iff the step K is coprime with N, so we want
    //   the largest K in [1, N/2] with gcd(N, K) = 1.
    //
    // - Start from K = floor(N/2) and walk downward, returning the first K that
    //   is coprime with N. Since N/2 and the next one or two values below it
    //   always include a value coprime with N, this stops after a couple of
    //   steps; arithmetic on the up-to-2000-digit N uses the big integer type.

    bigint k = n / 2;
    bigint one = 1;
    while(k > bigint(0)) {
        if(gcd(n, k) == one) {
            cout << k << '\n';
            break;
        }

        k -= one;
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}

=================
p193.in1
======================
7

=================
p193.in2
======================
6

=================
p193.py
======================
def gcd(a, b):
    if b == 0:
        return a
    return gcd(b, a % b)


n = int(input())

k = n // 2
while k > 0:
    if gcd(n, k) == 1:
        print(k)
        break

    k -= 1

=================
statement.txt
======================
193. Chinese Girls' Amusement
time limit per test: 0.25 sec.
memory limit per test: 65536 KB
input: standard
output: standard



You must have heard that the Chinese culture is quite different from that of Europe or Russia. So some Chinese habits seem quite unusual or even weird to us.

So it is known that there is one popular game of Chinese girls. N girls stand forming a circle and throw a ball to each other. First girl holding a ball throws it to the K-th girl on her left (1 ≤ K ≤ N/2). That girl catches the ball and in turn throws it to the K-th girl on her left, and so on. So the ball is passed from one girl to another until it comes back to the first girl. If for example N = 7 and K = 3, the girls receive the ball in the following order: 1, 4, 7, 3, 6, 2, 5, 1.

To make the game even more interesting the girls want to choose K as large as possible, but they want one condition to hold: each girl must own the ball during the game.

Input

Input file contains one integer number N (3 ≤ N ≤ 102000) - the number of Chinese girls taking part in the game.

Output

Output the only number - K that they should choose.

Sample test(s)

Input
Test #1
7

Test #2
6


Output
Test #1
3

Test #2
1

Author:	Alexander Maximov, Andrew Stankevich
Resource:	Petrozavodsk Winter Trainings 2003
Date:	2003-02-06





=================
