第几天？

#include <iostream>

using namespace std;

int s[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

int main() {
    int a, b, c;

    while (~scanf("%d/%d/%d", &a, &b, &c)) {
        if ((a % 4 == 0 && a % 100 != 0) || a % 400 == 0)
            s[1] = 29;
        else
            s[1] = 28;
        int ans = 0;

        for (int i = 0; i < b - 1; i ++ )

            ans += s[i];
        ans += c;
        cout << ans << endl;
    }

    return 0;
}

进制转换

#include <iostream>
#include <cstring>

using namespace std;

const int N = 1e5;

char m[N];

void solve(int a, int b) {
    int cnt = -1;
    memset(m, 0, sizeof (m));

    while (a) {
        m[++ cnt] = a % b < 10 ? (a % b + 48) : (a % b + 55);
        a /= b;
    }

    for (int i = cnt; i >= 0; i -- )

        cout << m[i];
    cout << endl;
}

int main() {
    int n, r;

    while (cin >> n >> r) {
        if (n >= 0) {
            solve(n, r);
        } else {
            cout << '-';
            solve(-n, r);
        }
    }

    return 0;
}

亲和数

#include <iostream>

using namespace std;

int solve(int x) {
    int ans = 0;

    for (int i = 1; i < x; i ++ )

        if (x % i == 0)
            ans += i;
    return ans;
}

int main() {
    int t;
    cin >> t;

    while (t -- ) {
        int a, b;
        cin >> a >> b;

        if (solve(a) == b && solve(b) == a)
            cout << "YES" << endl;
        else
            cout << "NO" << endl;
    }

    return 0;
}

一只小蜜蜂…

#include <iostream>

using namespace std;

typedef long long LL;

LL a[60];

void solve() {
    a[1] = 1;
    a[2] = 2;

    for (int i = 3; i < 60; i ++ )

        a[i] = a[i - 1] + a[i - 2];
}

int main() {
    solve();

    int t;

    while (cin >> t) {
        while (t -- ) {
            int n, m;
            cin >> n >> m;
            cout << a[m - n] << endl;
        }
    }



    return 0;
}

蟠桃记

#include <iostream>

using namespace std;

int main() {
    int n;

    while (cin >> n) {
        int ans = 1;

        for (int i = 1; i < n; i ++ ) {

            ans = (ans + 1) * 2;
        }

        cout << ans << endl;
    }

    return 0;
}