题目链接：http://lx.lanqiao.cn/problemset.page?code=BASIC

BASIC-1 闰年判断

题解

模拟。

代码

#include <bits/stdc++.h>
using namespace std;

int main() {
    int y;
    cin >> y;
    cout << ((y % 4 == 0 && y % 100 != 0) || (y % 400 == 0) ? "yes" : "no") << "\n";
    return 0;
}

BASIC-2 01字串

题解

五重循环枚举或DFS均可。

代码

#include <bits/stdc++.h>
using namespace std;

string s = "00000";

void dfs(int dep) {
    if (dep == 5) {
        cout << s << "\n";
        return;
    }
    s[dep] = '0';
    dfs(dep + 1);
    s[dep] = '1';
    dfs(dep + 1);
}

int main() {
    dfs(0);
    return 0;
}

BASIC-3 字母图形

题解

观察发现A的位置与所在行数相关，之后左右延伸出BCD...。

Tips

当行数比列数大时，枚举A时会越界，可以先预设出足够的空间。

代码

 

#include <bits/stdc++.h>
using namespace std;
const int N = 30;

int n, m;
vector<string> MP(N, string(N, '#'));

int main() {
    cin >> n >> m;
    for (int i = 0; i < n; i++) {
        char ch = 'A';
        for (int j = i; j < m; j++)
            MP[i][j] = ch++;
        ch = 'A';
        for (int j = i; j >= 0; j--)
            MP[i][j] = ch++;
        for (int j = 0; j < m; j++)
            cout << MP[i][j];
        cout << "\n";
    }
    return 0;
}

BASIC-4 数列特征

题解

模拟。

代码

#include <bits/stdc++.h>
using namespace std;

int main() {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++)
        cin >> a[i];
    cout << *max_element(a.begin(), a.end()) << "\n";
    cout << *min_element(a.begin(), a.end()) << "\n";
    cout << accumulate(a.begin(), a.end(), 0) << "\n";
    return 0;
}

BASIC-5 查找整数

题解

模拟。

代码

#include <bits/stdc++.h>
using namespace std;

int main() {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++)
        cin >> a[i];
    int x;
    cin >> x;
    int pos = find(a.begin(), a.end(), x) - a.begin();
    cout << (pos < n ? pos + 1 : -1) << "\n";
    return 0;
}

BASIC-6 杨辉三角形

题解

模拟。

代码

#include <bits/stdc++.h>
using namespace std;
const int N = 50;

int main() {
    int n;
    cin >> n;
    vector<vector<int> > f(N, vector<int>(N));
    for (int i = 1; i < N; i++)
        f[i][1] = f[i][i] = 1;
    for (int i = 3; i < N; i++) {
        for (int j = 2; j < i; j++) {
            f[i][j] = f[i - 1][j - 1] + f[i - 1][j];
        }
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= i; j++) {
            cout << f[i][j] << " \n"[j == i];
        }
    }
    return 0;
}

BASIC-7 特殊的数字

题解

模拟。

代码

#include <bits/stdc++.h>
#define pow3(n) (n * n * n)
using namespace std;

bool ok(int n) {
    int a = n / 100;
    int b = n / 10 % 10;
    int c = n % 10;
    return pow3(a) + pow3(b) + pow3(c) == n;
}

int main() {
    for (int i = 100; i <= 999; i++) {
        if (ok(i))
            cout << i << "\n";
    }
    return 0;
}

BASIC-8 回文数

题解

根据回文数的性质构造出前半部分即可，注意第一位不能为0。

代码

#include <bits/stdc++.h>
using namespace std;
int main() {
    for (int i = 1; i <= 9; i++) {
        for (int j = 0; j <= 9; j++) {
            cout << i << j << j << i << "\n";
        }
    }
    return 0;
}

BASIC-9 特殊回文数

题解

同上。

代码

#include <bits/stdc++.h>
using namespace std;
int main() {
    int n;
    cin >> n;

    for (int i = 1; i <= 9; i++) {
        for (int j = 0; j <= 9; j++) {
            for (int k = 0; k <= 9; k++) {
                if (i + j + k + j + i == n) {
                    cout << i << j << k << j << i << "\n";
                }
            }        
        }
    }

    for (int i = 1; i <= 9; i++) {
        for (int j = 0; j <= 9; j++) {
            for (int k = 0; k <= 9; k++) {
                if (i + j + k + k + j + i == n) {
                    cout << i << j << k << k << j << i << "\n";
                }
            }        
        }
    }

    return 0;
}

BASIC-10 十进制转十六进制

题解

模拟。

Tips

考虑到 $n = 0$ 的情况，用 do while() 。

代码

#include <bits/stdc++.h>
using namespace std;
int main() {
    int n;
    cin >> n;
    string s;
    do {
        int m = n % 16;
        if (m <= 9) s += '0' + m;
        else s += 'A' + m - 10;
        n /= 16;
    } while (n);
    reverse(s.begin(), s.end());
    cout << s << "\n";
    return 0;
}

BASIC-11 十六进制转十进制

题解

模拟。

代码

#include <bits/stdc++.h>
using namespace std;

long long hex_to_dec(const string &hex) {
    long long res = 0;
    for (int i = 0; i < hex.size(); i++) {
        res = res * 16 + (hex[i] <= '9' ? (hex[i] - '0') : (10 + hex[i] - 'A'));
    }
    return res;
}

int main() {
    string str;
    cin >> str;
    cout << hex_to_dec(str) << "\n";
    return 0;
}

BASIC-12 十六进制转八进制

题解

模拟。

代码

#include <bits/stdc++.h>
using namespace std;

string hex_to_bin(const string &hex) {
    string bin;
    for (int i = 0; i < hex.size(); i++) {
        if (hex[i] == '0') bin += "0000";
        if (hex[i] == '1') bin += "0001";
        if (hex[i] == '2') bin += "0010";
        if (hex[i] == '3') bin += "0011";
        if (hex[i] == '4') bin += "0100";
        if (hex[i] == '5') bin += "0101";
        if (hex[i] == '6') bin += "0110";
        if (hex[i] == '7') bin += "0111";
        if (hex[i] == '8') bin += "1000";
        if (hex[i] == '9') bin += "1001";
        if (hex[i] == 'A') bin += "1010";
        if (hex[i] == 'B') bin += "1011";
        if (hex[i] == 'C') bin += "1100";
        if (hex[i] == 'D') bin += "1101";
        if (hex[i] == 'E') bin += "1110";
        if (hex[i] == 'F') bin += "1111";
    }
    while (bin[0] == '0') bin.erase(0, 1);
    if (bin.size() % 3) bin = string(3 - bin.size() % 3, '0') + bin;
    return bin;
}

string bin_to_oct(const string &bin) {
    string oct;
    for (int i = 0; i < bin.size(); i += 3) {
        string s = bin.substr(i, 3);
        if (s == "000") oct += "0";
        if (s == "001") oct += "1";
        if (s == "010") oct += "2";
        if (s == "011") oct += "3";
        if (s == "100") oct += "4";
        if (s == "101") oct += "5";
        if (s == "110") oct += "6";
        if (s == "111") oct += "7";
    }
    return oct;
}

void solve() {
    string str;
    cin >> str;
    cout << bin_to_oct(hex_to_bin(str)) << "\n";
}

int main() {
    int t;
    cin >> t;
    while (t--) solve();
    return 0;
}

BASIC-13 数列排序

题解

模拟。

代码

#include <bits/stdc++.h>
using namespace std;
int main() {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++)
        cin >> a[i];
    sort(a.begin(), a.end());
    for (int i = 0; i < n; i++)
        cout << a[i] << ' ';
    return 0;
}