//10287 - Gifts in a Hexagonal Box (by Snail)
#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;

int main() {
    double s, r1, r2, r3, r4;
    cout << fixed << setprecision(10);
    r1 = sqrt(3.) / 2;
    r2 = 2*sqrt(3.) - 3;
    r3 = sqrt(3.) / 4;
    r4 = .6*sqrt(7.) - .7*sqrt(3.);
    while (cin >> s) {
        cout << r1*s << ' ' << r2*s << ' '
             << r3*s << ' ' << r4*s << endl;
    }
}

//10878 - Decode the tape (by snail)
#include <iostream>
#include <string>
using namespace std;

int main () {
    string s;
    while (getline(cin, s)) {
        while (getline(cin, s), s !="___________") {
            cout << char((s[2]=='o')*64 +
                (s[3]=='o')*32 + (s[4]=='o')*16 +
                (s[5]=='o')*8 + (s[7]=='o')*4 +
                (s[8]=='o')*2 + (s[9]=='o'));
        }
        cout << endl;
    }
}

//424 - Integer Inquiry (by Snail)
#include <iostream>
#include <string>
using namespace std;

int main () {
    string n1, n2;
    cin >> n1;
    while (cin >> n2, n2 != "0") {
        if (n1.size() < n2.size())              // 確定 n1 比 n2 長
            swap (n1, n2);
        n2.insert (0, n1.size()-n2.size(), '0');// 把數字補成一樣長
        int c = 0;                              // c(arry) -- 進位
        for (int i=n1.size()-1; i>=0; i--) {    // 從個位數開始算
            c += n1[i] + n2[i] - 96;
            n1[i] = c % 10 + '0';
            c /= 10;
        }
        if (c)                                  // 若還有進位
            n1.insert (0, 1, '1');              // 再加一位數
    }
    cout << n1 << endl;
}

//10127 - Ones (by Snail)
#include <iostream>
using namespace std;

int main () {
    int n, m, c;
    while (cin >> n) {
        m = 1;
        for (c=1; m%n; c++)                     // c(ount) -- 計算幾位數
            m = (m * 10 + 1) % n;               // m(ultiple) -- 倍數
        cout << c << endl;
    }
}

//10193 - All You Need Is Love (by Snail)
#include <iostream>
#include <string>
using namespace std;

int main () {
    int n, a, b, t, p=1;
    string s1, s2;
    cin >> n;
    while (n--) {
        cin >> s1 >> s2;
        a = strtol (s1.c_str(), NULL, 2);
        b = strtol (s2.c_str(), NULL, 2);
        while (b)                               // 輾轉相除法
            t = a%b, a = b, b = t;
        if (a == 1)                             // 互質
            cout << "Pair #" << p++ << ": Love is not all you need!\n";
        else
            cout << "Pair #" << p++ << ": All you need is love!\n";
    }
}

//10325 - The Lottery (by Snail)
#include <iostream>
using namespace std;

int N, M, r[16];                                // r(andom)[i]--M 個隨機數

int cnt(long long lcm, int sgn, int i) {
    if (lcm > N) return 0;                      // 避免 lcm 溢位
    if (i > M) return N / (int) lcm * sgn;      // sgn (sign) -- 正負號, +1 與 -1 變換
    int a = (int)lcm, b = r[i], t;
    while (b)                                   // 當 b==0 時
        t = a%b, a = b, b = t;                  // a 為 lcm 與 r[i] 的最大公因數
    return cnt(lcm,         sgn, i+1) +         // 偶數個 r 的 lcm 的倍數要加回來
           cnt(lcm*r[i]/a, -sgn, i+1);          // 奇數個 r 的 lcm 的倍數要扣掉
}

int main() {
    while (cin >> N >> M) {
        for (int i=1; i<=M; i++)
            cin >> r[i];
        cout << cnt(1, 1, 1) << endl;
    }
}

//10041 - Vito's large family (by Snail)
#include <iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;


int main () {
    int t, r, i, s, ac;
    long long d;
    cin >> t;
    while (t--) {
        int c[30001] = {};                      // c(ount)[s] -- s 出現次數
        cin >> r;
        for (i=0; i<r; i++)
            scanf("%d", &s), c[s]++;
        ac = 0;                                 // ac(cumulation) -- 累積人數
        for (i=0; ac+c[i] < (r+1)/2; i++)
            ac += c[i];
        d = 0;                                  // d(istance) -- 總距離
        for (s=0; s<=30000; s++)
            d += abs(s - i) * c[s];
        cout << d << ' ' << i << endl;
    }
}