*********************** 第 1 題 ***********************
//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;
}
}
*********************** 第 2 題 ***********************
//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;
}
}
*********************** 第 3 題 ***********************
//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;
}
*********************** 第 4 題 ***********************
//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;
}
}
*********************** 第 5 題 ***********************
//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";
}
}
*********************** 第 6 題 ***********************
//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;
}
}
*********************** 第 7 題 ***********************
//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;
}
}