Codeforces Round 751 (Div. 2)

比赛链接：https://codeforces.ml/contest/1602

A. Two Subsequences

直接找到最小的字符输出，然后把剩下的也输出

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

typedef long long ll;
typedef pair<int, int> pii;
typedef unsigned long long ull;
const ll inf = 1e18;
const int N = 1e5 + 10;
const int M = 1e6 + 10;
const double eps = 1e-8;
const int mod = 1e9 + 7;

#define fi first
#define se second
#define re register
#define lowbit (-x&x)

void solve() {
    int T; cin >> T; while (T--) {
        string s; cin >> s;
        vector<int> vis(100+1);
        int idx = 0;
        for (int i = 1; i < s.size(); i++) {
            if (s[i]< s[idx]) {
                idx = i;
            }
        }
        vis[idx] = 1;
        cout << s[idx] << ' ';
        for (int i = 0; i < s.size(); i++) {
            if (!vis[i]) cout << s[i];
        }
        cout << endl;
    }
}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
#ifdef ACM_LOCAL
    freopen("input", "r", stdin);
    freopen("output", "w", stdout);
#endif

#ifdef ACM_LOCAL
    auto start = clock();
#endif
    int t = 1;
//    cin >> t;
    while (t--)
        solve();
#ifdef ACM_LOCAL
    auto end = clock();
    cerr << "Run Time: " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
#endif
    return 0;
}

B. Divine Array

可以看出超过n次操作肯定会趋于不变，因此$n^2$暴力一下

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

typedef long long ll;
typedef pair<int, int> pii;
typedef unsigned long long ull;
const ll inf = 1e18;
const int N = 1e5 + 10;
const int M = 1e6 + 10;
const double eps = 1e-8;
const int mod = 1e9 + 7;

#define fi first
#define se second
#define re register
#define lowbit (-x&x)
int a[N];
int b[2005][2005];
void solve() {
    int T; cin >> T; while (T--) {
        int n; cin >> n;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                b[i][j] = 0;
            }
        }
        for (int i = 1; i <= n; i++) cin >> a[i], b[0][i] = a[i];
        for (int i = 1; i <= n; i++) {
            vector<int> cnt(n+1);
            for (int j = 1; j <= n; j++) {
                cnt[b[i-1][j]]++;
            }
            for (int j = 1; j <= n; j++) {
                b[i][j] = cnt[b[i-1][j]];
            }
        }
        int q; cin >> q; while (q--) {
            int x, k; cin >> x >> k;
            if (k > n) cout << b[n][x] << endl;
            else cout << b[k][x] << endl;
        }
    }
}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
#ifdef ACM_LOCAL
    freopen("input", "r", stdin);
    freopen("output", "w", stdout);
#endif

#ifdef ACM_LOCAL
    auto start = clock();
#endif
    int t = 1;
//    cin >> t;
    while (t--)
        solve();
#ifdef ACM_LOCAL
    auto end = clock();
    cerr << "Run Time: " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
#endif
    return 0;
}

C. Array Elimination

从进制角度考虑，每次取k个&肯定会消除k个1，因此每位上的和都是k的倍数，那么可以实现。

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

typedef long long ll;
typedef pair<int, int> pii;
typedef unsigned long long ull;
const ll inf = 1e18;
const int N = 2e5 + 10;
const int M = 1e6 + 10;
const double eps = 1e-8;
const int mod = 1e9 + 7;

#define fi first
#define se second
#define re register
#define lowbit (-x&x)
int a[N];
int sum[31];
void solve() {
    int T; cin >> T; while (T--) {
        int n;
        cin >> n;
        for (int i = 1; i <= n; i++) cin >> a[i];
        for (int i = 0; i < 30; i++) sum[i] = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j < 30; j++) {
                sum[j] += (a[i] >> j & 1);
            }
        }

        for (int i = 1; i <= n; i++) {
            int ff = 0;
            for (int j = 0; j < 30; j++) {
                if (sum[j] % i != 0) {
                    ff = 1;
                }
            }
            if (!ff) cout << i << ' ';
        }
        cout << endl;
    }
}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
#ifdef ACM_LOCAL
    freopen("input", "r", stdin);
    freopen("output", "w", stdout);
#endif

#ifdef ACM_LOCAL
    auto start = clock();
#endif
    int t = 1;
//    cin >> t;
    while (t--)
        solve();
#ifdef ACM_LOCAL
    auto end = clock();
    cerr << "Run Time: " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
#endif
    return 0;
}

D. Frog Traveler

有点毒瘤的题，开始还以为是线段树优化建图，然后跑最短路。赛中忽略了一个重要的条件就是不能往回跳，利用这个条件剪枝就可以了。

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

typedef long long ll;
typedef pair<int, int> pii;
typedef unsigned long long ull;
const ll inf = 1e18;
const int N = 3e5 + 10;
const int M = 1e6 + 10;
const double eps = 1e-8;
const int mod = 1e9 + 7;

#define fi first
#define se second
#define re register
#define lowbit (-x&x)
#define endl '\n'
int a[N], b[N], dis[N], n, pre[N];
void print(int x) {
    if (x == n) return;
    print(pre[x]);
    cout << x << ' ';
}
void solve() {
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1; i <= n; i++) cin >> b[i];
    memset(dis, 0x3f, sizeof dis);
    queue<int> q;
    q.push(n);
    int dep = n;
    dis[n] = 0;
    while (q.size()) {
        int now = q.front();
        q.pop();
        if (now <= 0) {
            cout << dis[now] << endl;
            print(now);
            return;
        }
        int x = now + b[now];
        for (int v = min(dep-1, x-1); v >= x-a[x]; v--) {
            if (dis[v] > dis[now] + 1) {
                dis[v] = dis[now] + 1;
                pre[v] = now;
                q.push(v);
            }
        }
        dep = min(dep, x - a[x]);
    }
    cout << -1 << endl;
}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
#ifdef ACM_LOCAL
    freopen("input", "r", stdin);
    freopen("output", "w", stdout);
#endif

#ifdef ACM_LOCAL
    auto start = clock();
#endif
    int t = 1;
//    cin >> t;
    while (t--)
        solve();
#ifdef ACM_LOCAL
    auto end = clock();
    cerr << "Run Time: " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
#endif
    return 0;
}