目录

A. X Axis

B. Matrix Stabilization

C. Update Queries

D. Mathematical Problem

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A. X Axis

Problem - A - Codeforces

直接找到第二大的数，答案就是这个数与其他两个数的差值的和。

void solve()
{
    vector<ll>a;
    for (int i = 1; i <= 3; i++)
    {
        int x;
        cin >> x;
        a.push_back(x);
    }
    sort(a.begin(), a.end());
    cout << a[2]-a[1]+a[1]-a[0] << "\n";
   
}

B. Matrix Stabilization

Problem - B - Codeforces

我们发现符合题意的每个数是互不影响的，即改变了其中一个不会影响到另一个，直接找到所以符合题意的数，再将这个数变成周围数的最大值。

int MX(int a, int b, int c, int d)
{
    vector<int>q;
    q.push_back(a);
    q.push_back(b);
    q.push_back(c);
    q.push_back(d);
    sort(q.begin(), q.end());
    return q[3];
}
void solve()
{
    memset(f, 0, sizeof(f));
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            cin >> f[i][j];
        }
    }
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            if (f[i][j] > f[i - 1][j] && f[i][j] > f[i][j - 1] && f[i][j] > f[i + 1][j] && f[i][j] > f[i][j + 1])
            {
                f[i][j] = MX(f[i - 1][j], f[i + 1][j], f[i][j - 1], f[i][j + 1]);
            }
        }
    }
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            cout<< f[i][j]<<" ";
        }
        cout << "\n";
    }
    cout << "\n";
}

C. Update Queries

Problem - C - Codeforces

仔细看就会发现这个操作数组顺序其实没多大关系，只要找到需要操作的所以，再一次填入已经排好序的元素即可。

ll d[200010];
void solve()
{
    memset(d, 0, sizeof(d));
    ll n, m;
    cin >> n >> m;
    string s1, s2;
    cin >> s1;
    int x;
    for (int i = 1; i <= m; i++) cin >> x, d[x]++;
    cin >> s2;
    sort(s2.begin(), s2.end());
    int c = 0;
    for (int i = 0; i < s1.size(); i++)
    {
        if (d[i + 1]&&c<s2.size())
        {
            s1[i] = s2[c++];
        }
    }
    cout << s1 << "\n";
}

D. Mathematical Problem

Problem - D - Codeforces

考虑n大于2的情况

当含有元素0时，那么答案就是0，因为可以将他们全部相乘。

当所以元素都为1时，答案就是1，可以全部相乘。

其他情况，答案就是不为1的所有元素的和。

void solve()
{
    int n;
    cin >> n;

    string s;
    cin >> s;

    vector<int> cnt(10);
    int sum = 0;
    for (int i = 1; i < n; i++) {
        if (s[i] <= '1') {
            cnt[s[i] - '0']++;
        }
        sum += s[i] - '0';
    }

    constexpr int inf = 1e9;
    int ans = inf;

    for (int i = 0; i + 2 <= n; i++) {
        cnt[s[i + 1] - '0']--;
        sum -= s[i + 1] - '0';

        int x = stoi(s.substr(i, 2));

        cout << x << ' ' << sum << ' ' << cnt[0] << ' ' << cnt[1] << '\n';

        int res;
        if (cnt[0] + (x == 0) > 0) {
            res = 0;
        }
        else if (cnt[1] + (x == 1) == n - 1) {
            res = 1;
        }
        else {
            res = (x != 1 ? x : 0) + sum - cnt[1];
        }
        ans = min(ans, res);

        cnt[s[i] - '0']++;
        sum += s[i] - '0';
    }

    cout << ans << '\n';
}