Codeforces Round 898 (Div. 4)

1|0A. Short Sort

#include<bits/stdc++.h>

using namespace std;


int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    int t;
    cin >> t;
    for( string s ; t ; t -- ){
        cin >> s;
        if( s == "abc" or s == "acb" or s == "cba" or s == "bac" )
            cout << "YES\n";
        else
            cout << "NO\n";
    }
    return 0;
}

2|0B. Good Kid

最小值加一

#include<bits/stdc++.h>

using namespace std;

#define int long long

void solve(){
    int n , res = 1 ;
    cin >> n;
    vector<int> a(n);
    for( auto &i : a )
        cin >> i;
    sort( a.begin(), a.end() );
    a[0] ++;
    for( auto i : a ) res *= i;
    cout << res << "\n";
    return;
}

int32_t main() {
    ios::sync_with_stdio(0), cin.tie(0);
    int t;
    for( cin >> t; t ; t -- )
        solve();

    return 0;
}

3|0C. Target Practice

#include<bits/stdc++.h>

using namespace std;

#define int long long

vector<string> p = {"1111111111",
                    "1222222221",
                    "1233333321",
                    "1234444321",
                    "1234554321",
                    "1234554321",
                    "1234444321",
                    "1233333321",
                    "1222222221",
                    "1111111111"};

void solve() {
    string s;
    int res = 0;
    for (int i = 0; i < 10; i++) {
        cin >> s;
        for (int j = 0; j < 10; j++) {
            if (s[j] == 'X')
                res += p[i][j] - '0';
        }
    }
    cout << res << "\n";
}

int32_t main() {
    ios::sync_with_stdio(0), cin.tie(0);
    int t;
    for (cin >> t; t; t--)
        solve();

    return 0;
}

4|0D. 1D Eraser

简单dp

#include<bits/stdc++.h>

using namespace std;

#define int long long

void solve() {
    int n, k;
    string s;
    cin >> n >> k >> s;
    vector<int> a(n + 1), f(n + 1, INT_MAX);
    for (int i = 1; i <= n; i++)
        a[i] = a[i - 1] + (s[i - 1] == 'B');
    f[0] = 0;
    for (int i = 1; i < k ; i++) {
        if (a[i] == 0) f[i] = 0;
        else f[i] = 1;
    }
    for (int i = k; i <= n; i++) {
        if (a[i] == a[i - 1]) f[i] = min(f[i], f[i - 1]);
        f[i] = min(f[i], f[i - k] + (a[i] > a[i - k]));
    }
    cout << f[n] << "\n";
    return;
}

int32_t main() {
    ios::sync_with_stdio(0), cin.tie(0);
    int t;
    for (cin >> t; t; t--)
        solve();

    return 0;
}

5|0E. Building an Aquarium

二分答案

#include<bits/stdc++.h>

using namespace std;

#define int long long


void solve() {
    int n, m;
    cin >> n >> m;
    vector<int> a(n);
    for (auto &i: a) cin >> i;

    auto check = [m, a](int h) {
        int cnt = 0;
        for (int i: a) cnt += max(0ll, h - i);
        return cnt <= m;
    };

    int l = 0, r = 2e9 + 5, res = 0;
    for (int mid; l <= r;) {
        mid = (l + r) / 2;
        if (check(mid)) res = mid, l = mid + 1;
        else r = mid - 1;
    }
    cout << res << "\n";
    return;
}

int32_t main() {
    ios::sync_with_stdio(0), cin.tie(0);
    int t;
    for (cin >> t; t; t--)
        solve();

    return 0;
}

6|0F. Money Trees

首先预处理每个点向后最远能到达的位置，然后前缀和维护一下果子的数量，最后二分长度即可。

#include<bits/stdc++.h>

using namespace std;

#define int long long


void solve() {
    int n, k;
    cin >> n >> k;
    vector<int> a(n + 1), h(n + 1), lst(n + 1);
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1; i <= n; i++) cin >> h[i];
    for (int i = 1; i <= n; i++) a[i] += a[i - 1];
    for (int i = 1, t; i <= n; i = t) {
        for (t = i + 1; t <= n; t++)
            if (h[t - 1] % h[t] != 0) break;
        for (int j = i; j < t; j++)
            lst[j] = t - 1;
    }

    auto check = [k,n,a,lst]( int x ){
        for( int l = 1 , r = x ; r <= n ; l ++ , r ++ ){
            if( r > lst[l] ) continue;
            if( a[r] - a[l-1] <= k ) return true;
        }
        return false;
    };

    int l = 1, r = n, res = 0;
    for (int mid; l <= r;) {
        mid = ( l + r ) / 2 ;
        if( check(mid) ) res = mid , l = mid + 1;
        else r = mid - 1;
    }
    cout << res << "\n";
    return;
}

int32_t main() {
    ios::sync_with_stdio(0), cin.tie(0);
    int t;
    for (cin >> t; t; t--)
        solve();
    return 0;
}

7|0G. ABBC or BACB

一个B可以把两侧的A全部消掉，所以只有当B的数量等于A的段数减一时，会有一段A无法被消掉，此时舍弃最小的一段。

#include <bits/stdc++.h>

using namespace std;

#define int long long

const int inf = 1e18;

using vi = vector<int>;

void solve() {
    string s;
    cin >> s , s += 'B';
    int cntB = -1, res;
    vector<int> a;
    for (int t = 0; auto i: s) {
        if (i == 'A') t++;
        else {
            cntB++;
            if (t > 0) a.push_back(t);
            t = 0;
        }
    }
    res = accumulate(a.begin(), a.end(), 0);
    if (cntB == a.size() - 1) {
        sort(a.begin(), a.end());
        res -= a.front();
    }
    cout << res << "\n";
    return;
}

int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);

    int t;
    for (cin >> t; t; t--)
        solve();

    return 0;
}

8|0H. Mad City

题目保证联通，所以图一定是基环树，要想实现永远追不上，这 b 要先跑到环上，所以a 要赶在b 到环上之前把他截住。所以找到 b 所在链和环的交点，判断谁先到交点即可。

#include<bits/stdc++.h>

using namespace std;

#define int long long

using vi = vector<int>;

constexpr int inf = 1e9;


void solve() {
    int n, a, b;
    cin >> n >> a >> b;
    vector<vi> e(n + 1);
    vi deg(n + 1), vis(n + 1);
    for (int i = 1, u, v; i <= n; i++) {
        cin >> u >> v;
        e[u].push_back(v), e[v].push_back(u);
        deg[u]++, deg[v]++;
    }
    queue<int> q;
    for (int i = 1; i <= n; i++)
        if (deg[i] == 1) q.push(i), vis[i] = 1;
    for (int u; !q.empty();) {
        u = q.front(), q.pop();
        for (auto v: e[u]) {
            if (vis[v]) continue;
            if (--deg[v] == 1) vis[v] = 1, q.push(v);
        }
    }

    auto bfs = [e, n](int u) {
        vi dis(n + 1, inf);
        queue<int> q;
        q.push(u), dis[u] = 0;
        for (int x; !q.empty();) {
            x = q.front(), q.pop();
            for (auto y: e[x]) {
                if (dis[y] <= dis[x] + 1) continue;
                dis[y] = dis[x] + 1, q.push(y);
            }
        }
        return dis;
    };
    vi d = bfs(b);
    int t = -1;
    for (int i = 1; i <= n; i++) {
        if (vis[i]) continue;
        if (t == -1 or d[i] < d[t]) t = i;
    }
    d = bfs(t);
    if( d[a] <= d[b] ) cout << "NO\n";
    else cout << "YES\n";
    return;
}

int32_t main() {
    ios::sync_with_stdio(0), cin.tie(0);
    int t;
    for (cin >> t; t; t--)
        solve();
    return 0;
}

__EOF__

本文作者：PHarr
本文链接：https://www.cnblogs.com/PHarr/p/17728882.html
关于博主：前OIer，SMUer
版权声明：CC BY-NC 4.0
声援博主：如果这篇文章对您有帮助，不妨给我点个赞