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弹跳

#include <bits/stdc++.h>
using namespace std;
long long x, y, ans;
int main()
{
    freopen("jump.in","r",stdin);
    freopen("jump.out","w",stdout);
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> x >> y;
    ans = x;
    while (x)
    {
        ans += x / y * 2;
        x /= y;
    }
    cout << ans;
    return 0;
}

闰年天数求和

#include <bits/stdc++.h>
using namespace std;
int l, r, ans;
bool leap(int y)
{
    if (y % 4 == 0 && y % 100 != 0 || y % 400 == 0)
        return true;
    return false;
}
int main()
{
    freopen("leap.in","r",stdin);
    freopen("leap.out","w",stdout);
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> l >> r;
    ans = 0;
    for (int i = l; i <= r; i++)
        if (leap(i))
            ans += 366;
    cout << ans;
    return 0;
}

文件IO检测

#include <bits/stdc++.h>
using namespace std;
string s, a, b, A, B;
int main()
{
    freopen("file.in","r",stdin);
    freopen("file.out","w",stdout);
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> s >> a >> b;
    A = "freopen(\"" + s + ".in\",\"r\",stdin);";
    B = "freopen(\"" + s + ".out\",\"w\",stdout);";
    if (a == A && b == B)
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

单点修改区间查询青春版

#include <bits/stdc++.h>
#define int long long
using namespace std;
int n, q;
int a[100000 + 5];
int sum[100000 + 5];
void initSum()
{
    sum[0] = 0;
    for (int i = 1; i <= n; i++)
        sum[i] = sum[i - 1] + a[i];
}
signed main()
{
    freopen("sum.in", "r", stdin);
    freopen("sum.out", "w", stdout);
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> q;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    initSum();
    while (q--)
    {
        int op, l, r, x, y;
        cin >> op;
        if (op == 1)
        {
            cin >> l >> r;
            cout << sum[r] - sum[l - 1] << "\n";
        }
        if (op == 2)
        {
            cin >> x >> y;
            a[x] = y;
            initSum();
        }
    }
    return 0;
}

走走走

https://www.luogu.com.cn/problem/CF2041D

7 15
###############
#S......#E....#
#.###########.#
#.......#.#...#
#######.#.....#
#..#......#...#
###############

30

30 分

#include <bits/stdc++.h>
using namespace std;
int n, m;
string s;
int main()
{
    freopen("walk.in", "r", stdin);
    freopen("walk.out", "w", stdout);
    cin >> n >> m;
    cin >> s;
    int sPos, ePos;
    for (int i = 0; i < s.size(); i++)
        if (s[i] == 'S')
            sPos = i;
        else if (s[i] == 'E')
            ePos = i;
    for (int i = sPos + 1; i <= ePos - 1; i++)
        if (s[i] == '#')
        {
            cout << "-1";
            return 0;
        }
    // 转换为 1->len
    int len = ePos - sPos + 1;
    int now = 1; // 当前位置
    int cnt = 0; // 连续向右的步骤
    int ans = 0; // 走的步数
    while (now != len)
    {
        if (cnt != 3)
        {
            // 往前走
            now++;
            cnt++;
            ans++;
        }
        else
        {
            // 后退
            now--;
            cnt = 0;
            ans++;
        }
    }
    cout << ans << "\n";
    return 0;
}

满分

#include <bits/stdc++.h>
using namespace std;
int n, m;
// 获取某个二维点和一维点的对应
int getIdx(int x, int y)
{
    return (x - 1) * m + y;
}
int getX(int idx)
{
    // idx/m 上取整
    return (idx + (m - 1)) / m;
}
int getY(int idx)
{
    // 转换到 0 开头的体系，% 完后加回来
    return (idx - 1) % m + 1;
}
// 地图与四个方向
char s[100000 + 5];
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
// 走到 i 这个位置，上一步在 j，沿着 j 方向连续走了 k 步，目前一共走了几步
int dis[100000 + 5][4][4];
struct Node
{
    int idx; // 点的编号
    int fx;  // 上一步方向
    int len; // fx 方向目前连续走的步数
};
queue<Node> q;
int main()
{
    freopen("walk.in","r",stdin);
    freopen("walk.out","w",stdout);
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> m;
    int sPos, ePos;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            int idx = getIdx(i, j);
            cin >> s[idx];
            if (s[idx] == 'S')
                sPos = idx;
            if (s[idx] == 'E')
                ePos = idx;
        }
    }
    // 起点入队，看作是 0 方向连续走了 0 步到达
    // 步数从 1 开始，这样步数为 0 的就是没走过的
    dis[sPos][0][0] = 1;
    q.push({sPos, 0, 0});
    while (!q.empty())
    {
        Node now = q.front();
        q.pop();
        int x = getX(now.idx); // 当前点
        int y = getY(now.idx); // 当前点
        int fx = now.fx;       // 当前点
        int len = now.len;     // 当前点
        if (now.idx == ePos)
        {
            cout << dis[now.idx][fx][len] - 1;
            return 0;
        }
        for (int i = 0; i < 4; i++)
        {
            int nx = x + dx[i];                   // 下一个点的属性
            int ny = y + dy[i];                   // 下一个点的属性
            int nIdx = getIdx(nx, ny);            // 下一个点的属性
            int nfx = i;                          // 下一个点的属性
            int nlen = (i == fx) ? (len + 1) : 1; // 下一个点的属性
            // 方向一样就多了一步，否则就是第一步
            if (nx < 1 || nx > n || ny < 1 || ny > m ||
                nlen == 4 ||
                dis[nIdx][nfx][nlen] != 0 ||
                s[nIdx] == '#')
                continue;
            dis[nIdx][nfx][nlen] = dis[now.idx][now.fx][now.len] + 1;
            q.push({nIdx, nfx, nlen});
        }
    }
    cout << -1 << "\n";
    return 0;
}

三四五

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

void work()
{
    long long a, aa;
    cin >> a;
    aa = a;
    if (a <= 2)
    {
        // 特判
        cout << -1 << "\n";
        return;
    }
    if (a % 2 == 1)
    {
        // 判断奇数
        a *= a;
        cout << aa << " ";
        cout << (a - 1) / 2 << " ";
        cout << (a + 1) / 2 << "\n";
    }
    if (a % 2 == 0)
    {
        // 判断偶数
        int ans = 1;
        while (a % 2 == 0)
        {
            a /= 2;
            ans *= 2;
        }
        if (a == 1) // 特判二次幂
        {
            cout << 3 * ans / 4 << " ";
            cout << aa << " ";
            cout << 5 * ans / 4 << "\n";
            return;
        }
        a *= a;
        cout << aa << " ";
        cout << (a - 1) * ans / 2 << " ";
        cout << (a + 1) * ans / 2 << "\n";
    }
}
int main()
{
    freopen("three.in", "r", stdin);
    freopen("three.out", "w", stdout);
    int n;
    cin >> n;
    while (n--)
        work();
    return 0;
}

玩游戏

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200000;
int n, m;
set<int> e[MAXN + 5]; // set 存图，方便删点
// 类似拓扑排序不停删入度为 2 的点的队列
int d[MAXN + 5];
queue<int> q;
// 还原时用到的链表
int nxt[MAXN + 5], pre[MAXN + 5];
bool vis[MAXN + 5];
void dfs(int u)
{
    cout << u << " ";
    vis[u] = true;
    int x = nxt[u];
    int y = pre[u];
    if (x > y)
        swap(x, y);
    if (!vis[x])
        dfs(x);
    if (!vis[y])
        dfs(y);
}
int main()
{
    freopen("circle.in", "r", stdin);
    freopen("circle.out", "w", stdout);
    ios::sync_with_stdio(false);
    cin.tie(0);
    // 存图
    cin >> n >> m;
    for (int i = 0; i < m; i++)
    {
        int u, v;
        cin >> u >> v;
        if (u == v)
            continue;
        e[u].insert(v);
        e[v].insert(u);
    }
    // 特判
    if (n <= 2)
    {
        for (int i = 1; i <= n; i++)
            cout << i << " ";
        return 0;
    }
    // 计算度数并把入度为 2 的入队
    for (int i = 1; i <= n; i++)
    {
        d[i] = e[i].size();
        if (d[i] == 2)
            q.push(i);
    }
    // 按顺序所有删除的点以及左右点
    // <度数为 2 的点, <左边的点, 右边的点>>
    vector<pair<int, pair<int, int>>> all2;
    int nn = n; // 剩余点数，超过两个点就丢一个入度为 2 的
    while (nn > 2)
    {
        int now = q.front();
        q.pop();
        if (e[now].size() != 2)
            continue;
        // 取出 now 相连的两个点
        int nowL = *(e[now].begin());
        e[now].erase(e[now].begin());
        int nowR = *(e[now].begin());
        e[now].erase(e[now].begin());
        // 记住 now 的左右点
        all2.push_back({now, {nowL, nowR}});
        // 在 nowL,nowR 中移除 now
        e[nowL].erase(now);
        e[nowR].erase(now);
        d[nowL]--;
        d[nowR]--;
        // 删除 now
        e[now].clear();
        d[now] = 0;
        nn--;
        // nowL与nowR 如果之前不相连就连边
        if (nowL != nowR && e[nowL].find(nowR) == e[nowL].end())
        {
            e[nowL].insert(nowR);
            e[nowR].insert(nowL);
            d[nowL]++;
            d[nowR]++;
        }
        // 检测 nowL,nowR 度数是否删到了 2
        if (d[nowL] == 2)
            q.push(nowL);
        if (d[nowR] == 2)
            q.push(nowR);
    }
    // 从剩余的两个点开始构造方案
    int u = -1, v = -1;
    for (int i = 1; i <= n; i++)
        if (d[i] != 0)
        {
            if (u == -1)
                u = i;
            else
                v = i;
        }
    nxt[u] = v, pre[u] = v;
    nxt[v] = u, pre[v] = u;
    for (int i = (int)all2.size() - 1; i >= 0; --i)
    {
        int now = all2[i].first;
        int nowL = all2[i].second.first;
        int nowR = all2[i].second.second;
        if (nxt[nowR] == nowL)
            swap(nowL, nowR);
        // 保证是：nowL ~ nowR
        // 下一步：nowL ~ now ~ nowR
        if (nxt[nowL] == nowR)
        {
            nxt[now] = nowR, pre[now] = nowL;
            nxt[nowL] = now, pre[nowR] = now;
        }
    }
    dfs(1);
    return 0;
}