AtCoder Beginner Contest 314

A - 3.14

#include <bits/stdc++.h>

using namespace std;

#define int long long




int32_t main() {
    ios::sync_with_stdio(0), cin.tie(0);
    string s = "1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679";
    int n;
    cin >> n;
    cout << "3.";
    for( int i = 0 ; i < n ; i ++ )
        cout << s[i];

    return 0;
}

B - Roulette

#include <bits/stdc++.h>

using namespace std;

#define int long long

int32_t main() {
    ios::sync_with_stdio(0), cin.tie(0);
    int n;
    cin >> n;
    vector<int> c(n+1);
    vector<set<int>> a(n+1);
    for( int i = 1 ; i <= n ; i ++ ){
        cin >> c[i];
        for( int j = 1 , x; j <= c[i] ; j ++ )
            cin >> x , a[i].insert(x);
    }
    int m;
    cin >> m;
    vector<int> res;
    for( int i = 1 ; i <= n ; i ++ ){
        if( a[i].count(m) ){
            if( res.empty() or c[res.back()] == c[i] ) res.push_back(i);
            else if( !res.empty() and c[res.back()] > c[i] ){
                res = vector<int>();
                res.push_back(i);
            }
        }
    }
    cout << res.size() << "\n";
    for( auto i : res )
        cout << i << " ";

    return 0;
}

C - Rotate Colored Subsequence

#include <bits/stdc++.h>

using namespace std;

#define int long long

int32_t main() {
    ios::sync_with_stdio(0), cin.tie(0);
    int n , m;
    cin >> n >> m;
    string s;
    cin >> s;
    vector<int> c( n );
    for( auto & i : c ) cin >> i;
    map<int , deque<char>> cnt;
    for( int i = 0 ; i < n ; i ++ )
        cnt[ c[i] ].push_back( s[i] );
    for( auto & [ k , q ] : cnt ){
        char c = q.back();
        q.pop_back() , q.push_front(c);
    }
    for( auto i : c ){
        cout << cnt[i].front();
        cnt[i].pop_front();
    }

    return 0;
}

D - LOWER

看起来需要很多数据结构。但实际上，操作 2、3 只有最后一次操作是有效的，所以记录最后一次操作位置即可。

#include <bits/stdc++.h>

using namespace std;

#define int long long


int32_t main() {
    ios::sync_with_stdio(0), cin.tie(0);
    int n;
    string s;
    cin >> n >> s;
    int q;
    cin >> q;
    vector<int> op(q), x(q);
    vector<char> c(q);
    int lst = -1;
    for (int i = 0; i < q; i++) {
        cin >> op[i] >> x[i] >> c[i];
        if (op[i] != 1) lst = i;

    }
    for (int i = 0; i < lst; i++) {
        if (op[i] != 1) continue;
        s[x[i]-1] = c[i];
    }
    if (lst >= 0 and op[lst] == 2) {

        for (auto &i: s)
            if (i >= 'A' && i <= 'Z') i = i + 'a' - 'A';
    } else if (lst >= 0 and op[lst] == 3) {
        for (auto &i: s)
            if (i >= 'a' && i <= 'z') i = i + 'A' - 'a';
    }
    for (int i = lst + 1; i < q; i++)
        s[x[i]-1] = c[i];
    cout << s;
    return 0;
}