Codeforces Beta Round #69(Div 2)

Apr 19, 2011.

Accepted: 3.        WA: 0.        NotTried: 2.

Tutorial

Problem 1:

Solution:

Could check every number from n+1,
if that number is a prime and also equals to m, then "YES".
else "NO";

Source Code:

//Tuesday, April  19, 2011 09:37 CDT
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

using namespace std;

bool isPrime(int n) {
    if (n % 2 == 0 && n != 2)
        return false;
    if (n == 2)
        return true;
    for (int i = 3; i * i <= n; i += 2)
        if (n % i == 0)
            return false;
    return true;
}

bool check(int n, int m) {
    int i;
    for (i = n + 1;; i++)
        if (isPrime(i))
            break;
    return i == m;
}

int main(int argc, char* argv[]) {
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    int n, m;
    cin >> n >> m;
    if (check(n, m))
        cout << "YES" << endl;
    else
        cout << "NO" << endl;

    //fclose(stdin);
    //fclose(stdout);
    return 0;
}

Problem 2:

Solution:

Easy Math problem, remember to check hour to make sure that's smaller than 12.

Source Code:

//Tuesday, April  19, 2011 09:37 CDT
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

using namespace std;

pair<double, double> solve(int h, int m) {
    h %= 12;
    double mm = 6.0 * m;
    double hh = 30.0 * h + 1.0 * m * 30.0 / 60.0;
    return make_pair(hh, mm);
}

int main(int argc, char* argv[]) {
    //  freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    int h, m;
    char c;
    cin >> h >> c >> m;
    pair<double, double> display = solve(h, m);
    cout << display.first << " " << display.second << endl;
    //  fclose(stdin);
    //fclose(stdout);
    return 0;
}

Problem 3:

Solution:

Source Code:

Problem 4:

Solution:

Source Code:

//Tuesday, April  19, 2011 09:37 CDT
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

using namespace std;

double solve(double a, double b) {
    if (b == 0)
        return 1.0;
    if (a == 0)
        return 0.5;
    if (4.0 * b > a) {
        return (0.5 + a / b / 16.0);
    } else
        return (1.0 - b / a);
}

int main(int argc, char* argv[]) {
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    int T;
    cin >> T;
    while (T--) {
        double a, b;
        cin >> a >> b;
        cout << solve(a, b) << endl;
    }
    //fclose(stdin);
    //fclose(stdout);
    return 0;
}

Problem 5:

Solution:

Source Code: