abc372

A

signed main() {
    IOS;
    string a;
    cin >> a;
    for (int i = 0; i < a.size(); i++) {
        if (a[i] != '.') {
            cout << a[i];
        }
    }

        return 0;
}

B

普普通通的三进制

signed main() {
    IOS;
    int x;
    cin >> x;
    int tmp = x;
    int tot = 0;
    int res = 0;
    while (x)
    {
        int idx = x % 3;
        while (idx--)
        {
            res++;
        }
        x /= 3;
    }
    cout << res << endl;
    x = tmp;
    while (x)
    {
        int idx = x % 3;
        while (idx--) {
            cout << tot << ' ';
        }
        tot++;
        x /= 3;
    }

    return 0;
}

C

随随便便的写一写

signed main() {
    IOS;
    int len, q;
    cin >> len >> q;
    string s;
    cin >> s;
    int ans = 0;
    for (int i = 0; i < s.size() - 2; i++)
    {
        if (s[i] == 'A' && s[i + 1] == 'B' && s[i + 2] == 'C') {
            ans++;
        }
    }
    while (q--) {
        int x;
        cin >> x;
        char c;
        cin >> c;
        ll tmp = 0;
        x -= 1;
        for (int i = max(x - 2, 0); i <= x; i++)
        {
            if (s[i] == 'A' && s[i + 1] == 'B' && s[i + 2] == 'C')
            {
                tmp++;
            }
        }
        s[x] = c;
        for (int i = max(x - 2, 0); i <= x; i++)
        {
            if (s[i] == 'A' && s[i + 1] == 'B' && s[i + 2] == 'C')
            {
                tmp--;
            }
        }
        ans -= tmp;
        cout << ans << endl;
    }

        return 0;
}

D

单调栈的板子

signed main()
{
    IOS;
    int n;
    cin >> n;
    deque<int> q;
    q.push_back(0);
    a[0] = 1e9;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
        while (!q.empty() && a[q.back()] < a[i])
        {
            q.pop_back();
        }
        b[q.back()]++, b[i]--;
        // cout << q.back() << ' ' << i << endl;
        q.push_back(i);
    }
    for (int i = 1; i <= n; i++)
    {
        // cout << b[i] << ' ';
        b[i] += b[i - 1];
    }
    // cout << endl;
    for (int i = 1; i <= n; i++)
    {
        cout << b[i] << ' ';
    }

    return 0;
}

E

很奇怪的并查集

#define MAX 200005
vector<int> G[MAX];
int fa[MAX];

int finds(int x) {
    return x == fa[x] ? x : fa[x] = finds(fa[x]);
}

void merge(int x, int y) {
    int nx = finds(x), ny = finds(y);
    if (nx == ny) {
        return;
    }
    vector<int> tmp;
    int tot = 0;
    for (int i = 0, j = 0; i < G[nx].size() || j < G[ny].size(); )
    {
        if (tot > 10) {
            break;
        }
        tot++;
        if (i < G[nx].size() && j < G[ny].size())
        {
            if (G[nx][i] > G[ny][j]) {
                tmp.push_back(G[nx][i]);
                i++;
            } else {
                tmp.push_back(G[ny][j]);
                j++;
            }
            continue;
            }
        if (i < G[nx].size()) {
            tmp.push_back(G[nx][i]);
            i++;
        } else {
            tmp.push_back(G[ny][j]);
            j++;
        }
    }
    G[nx] = tmp;
    fa[ny] = nx;
    G[ny].clear();
}

signed main()
{
    IOS;
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
        fa[i] = i;
        G[i].push_back(i);
    }
    while (m--) {
        int op, x, y;
        cin >> op >> x >> y;
        if (op == 1) {
            merge(x, y);
        } else {
            int nx = finds(x);
            if (G[nx].size() < y) {
                cout << -1 << endl;
            } else {
                cout << G[nx][y - 1] << endl;
            }
        }
    }

        return 0;
}

F

来回转就好了

const int MAXN=2e5;
int n,m,k;
pair<long long,long long>E[52];
long long dp[2*MAXN+10],tmp[52];
int main(){
    cin>>n>>m>>k;m++;
    E[1]=(make_pair(n,1));
    for(int i=2;i<=m;i++){
        cin>>E[i].first>>E[i].second;
    }
    dp[k]=1;
    for(int i=k-1;i>=0;i--){
        for(int j=1;j<=m;j++){
            tmp[j]=dp[i+E[j].first];
        }
        for(int j=1;j<=m;j++){
//            cout<<i+E[j].second-1<<' '<<i+E[j].first<<'\n';
            dp[i+E[j].second-1]=(dp[i+E[j].second-1]+tmp[j])%mod;
        }
//        cout<<"***\n";
    }
    long long ans=0;
    for(int i=0;i<n;i++){
        ans=(ans+dp[i])%mod;
    }
    cout<<ans;
    return 0;
}