题目见这

试题C 数列求值

#include <iostream>

using namespace std;

char str[27] = {'0', 'A', 'B', 'C', 'D', 'E', 'F', 
                'G', 'H', 'I', 'J', 'K', 'L', 'M', 
                'N', 'O', 'P', 'Q','R', 'S', 'T',
                'U', 'V','W', 'X', 'Y', 'Z'};

int main() {
    int num = 2019;
    string ans = "";
    while (num) {
        ans += str[num % 26];
        num /= 26;
    }
    for (int i = ans.size() - 1; i >= 0; i--) {
        printf("%c", ans[i]);
    }
    return 0;
}

试题D 数的分解

#include <cstdio>

using namespace std;
typedef long long LL;

bool st[2019];

int main() {
    int res = 0;
    for (int i = 2; i <= 2019; i++) {
        int t = i;
        while (t) {
            int k = t % 10;
            if (k == 2 || k == 4) {
                st[i] = true;
                break;
            }
            t /= 10;
        }
    }

    for (int i = 1; i < 2019; i++)
        for (int j = i + 1; j < 2019; j++) {
            int k = 2019 - i - j;
            if (k == 0 || k <= i || k <= j) continue;
            if (st[i] || st[j] || st[k]) continue;
            res++;
        }

    printf("%d\n", res);
    return 0;
}

试题E 迷宫

#include <iostream>
#include <cstring>
#include <queue>

using namespace std;
const int N = 50;
char g[N][N];
bool st[N][N];
int n, m;

struct Node {
    int x;
    int y;
    string path;
};

int dx[4] = {1, 0, 0, -1}, dy[4] = {0, -1, 1, 0};
char dir[4] = {'D', 'L', 'R', 'U'};

int main() {
    scanf("%d%d", &n, &m);

    for (int i = 1; i <= n; i++) scanf("%s", g[i] + 1);

    queue<Node> q;
    q.push({1, 1, ""});
    st[1][1] = true;

    while (!q.empty()) {
        Node t = q.front();
        q.pop();
        if (t.x == n && t.y == m) {
            cout << t.path;
            break;
        }
        for (int i = 0; i < 4; i++) {
            int a = t.x + dx[i], b = t.y + dy[i];
            if (a <= 0 || b <= 0 || a > n || b > m) continue;
            if (st[a][b]) continue;
            if (g[a][b] == '1') continue;
            q.push({a, b, t.path + dir[i]});
            st[a][b] = true;
        }
    }

    return 0;
}

试题F 特别数的和

#include <cstdio>

using namespace std;
int n;

int main() {
    scanf("%d", &n);
    int res = 0;
    for (int i = 1; i <= n; i++) {
        int t = i;
        while (t) {
            int k = t % 10;
            if (k == 2 || k == 0 || k == 1 || k == 9) {
                res += i;
                break;
            }
            t /= 10;
        }
    }

    printf("%d\n", res);
    return 0;
}

试题G 完全二叉树的权值

#include <iostream>

using namespace std;
const int N = 100010;
int a[N];
int n;

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);

    int len = 1;
    int l = 1, r = 1;
    int max_line = 0;
    int line = 1;
    int res = 0;
    while (true) {
        int sum = 0;
        int i = l;
        for (; i <= min(r, n); i++) sum += a[i];
        if (res < sum) {
            res = max(res, sum);
            max_line = line;
        }
        if (i > n) break;
        len *= 2;
        line ++;
        l = r + 1;
        r = r + len;
    }
    printf("%d\n", max_line);
    return 0;
}

试题H 等差数列

#include <iostream>
#include <algorithm>

using namespace std;

const int N = 100010;
int a[N];
int n;

int gcd(int a, int b) {
    return b ? gcd(b, a % b): a;
}

int main() {
    scanf("%d", &n);
    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    sort(a, a + n);
    int min_gcd = 0;
    for (int i = 1; i < n; i++) {
        min_gcd = gcd(min_gcd, a[i] - a[0]);
        if (!min_gcd) {
            printf("%d", n);
            return 0;
        }
    }
    printf("%d\n", (a[n - 1] - a[0]) / min_gcd + 1);

    return 0;
}

试题I 后缀表达式

#include <iostream>
#include <algorithm>

using namespace std;
typedef long long LL;

const int N = 100010;
int n, m;
int a[N + N];

int main() {
    scanf("%d%d", &n, &m);
    int num = n + m + 1;
    for (int i = 0; i < num; i++) scanf("%d", &a[i]);
    LL res = 0;
    if (!m) {
        for (int i = 0; i < num; i++) 
            res += a[i];
    } else {
        sort(a, a + num);
        res = res + a[num - 1] - a[0];
        for (int i = 1; i < num - 1; i++) 
            res += abs(a[i]);
    }

    printf("%lld\n", res);
    return 0;
}

试题J 灵能传输

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long LL;

const int N = 300010;

int n;
LL a[N], s[N];
bool st[N];

int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) {
            scanf("%lld", &a[i]);
            s[i] = s[i - 1] + a[i];
        }

        LL s0 = s[0], sn = s[n];
        if (s0 > sn) swap(s0, sn);
        sort(s, s + n + 1);

        for (int i = 0; i <= n; i++)
            if (s[i] == s0) {
                s0 = i;
                break;
            }

        for (int i = n; i >= 0; i--)
            if (s[i] == sn) {
                sn = i;
                break;
            }

        memset(st, 0, sizeof st);
        int l = 0, r = n;
        for (int i = s0; i >= 0; i -= 2) {
            a[l++] = s[i];
            st[i] = true;
        }
        for (int i = sn; i <= n; i += 2) {
            a[r--] = s[i];
            st[i] = true;
        }
        for (int i = 0; i <= n; i++)
            if (!st[i])
                a[l++] = s[i];

        LL res = 0;
        for (int i = 1; i <= n; i++) res = max(res, abs(a[i] - a[i - 1]));

        printf("%lld\n", res);
    }

    return 0;
}