Educational Codeforces Round 65 (Rated for Div. 2)

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于 2019-09-22 18:54:03 发布

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分类专栏： CodeForces

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本文精选了四道算法竞赛题目，包括电话号码合法性判断、丢失数字查找、新闻分发问题和双色RBS染色问题的解决方案。通过C++代码详细解析了每道题目的解题思路和实现细节。

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A. Telephone Number

思路：字符串长度小于11或长度为11时首位不为8或长度>11且第一个8的位置不<=n-11的情况都为No

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 3e5 + 5;
int main() {
    ios::sync_with_stdio(false);
    int T;
    cin >> T;
    while(T--) {
        int n;
        cin >> n;
        string str;
        cin >> str;
        if(n < 11)
            cout << "NO" << endl;
        else if(n == 11) {
            if(str[0] == '8')
                cout << "YES" << endl;
            else
                cout << "NO" << endl;
        } else {
            int x = str.size();
            for(int i = 0; i < str.size(); i += 1) {
                if(str[i] == '8') {
                    x = i;
                    break;
                }
            }
            if(x <= n - 11)
                cout << "YES" << endl;
            else
                cout << "NO" << endl;
        }
    }
    return 0;
}

B. Lost Numbers

按题目的意思写？

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
vector<int> p = {4, 8, 15, 16, 23, 42};
int query[4];
int main() {
    for(int i = 0; i < 4; i += 1) {
        cout << "? " << i + 1 << ' ' << i + 2 << endl;
        cout.flush();
        cin >> query[i];
    }
    do {
        bool yes = true;
        for(int i = 0; i < 4; i += 1) {
            if(p[i]*p[i + 1] != query[i]) {
                yes = false;
                break;
            }
        }
        if(yes)
            break;
    } while(next_permutation(p.begin(), p.end()));
    cout << "!";
    for(int i = 0; i < 6; i += 1)
        cout << ' ' << p[i];
    cout << '\n';
    cout.flush();
    return 0;
}

C. News Distribution

并查集基础题

#include<cstdio>
#include<algorithm>
using namespace std;
const int N = 5e5 + 5;
int fa[N], _rank[N], p[N];
int find(int x) {
    if(x == fa[x])
        return x;
    else
        return fa[x] = find(fa[x]);
}
void Merge(int x, int y) {
    int u = find(x);
    int v = find(y);
    if(u != v) {
        fa[v] = u;
        _rank[u] += _rank[v];
        _rank[v] = 0;
    }
}
void init(int n) {
    for(int i = 1; i <= n; i += 1) {
        fa[i] = i;
        _rank[i] = 1;
    }
}
int main() {
    int n, group;
    scanf("%d%d", &n, &group);
    init(n);
    while(group--) {
        int k;
        scanf("%d", &k);
        for(int i = 0; i < k; i += 1) {
            scanf("%d", &p[i]);
            Merge(p[0], p[i]);
        }
    }
    for(int i = 1; i <= n; i += 1)
        printf("%d%c", _rank[find(i)], i < n ? ' ' : '\n');
    return 0;
}

D. Bicolored RBS
思路：以相邻的()进行轮流01染色直到染完即可

#include<bits/stdc++.h>
using namespace std;
int main() {
    int n;
    string str;
    cin >> n >> str;
    stack<int> s;
    vector<int> v(n);
    v[0] = 0;
    s.push(0);
    for(int i = 1; i < n; ++i) {
        if(str[i] == '(') {
            if(s.empty()) {
                s.push(0);
                v[i] = 0;
            } else {
                int x = s.top();
                s.push(x ^ 1);
                v[i] = (x ^ 1);
            }
        } else {
            int x = s.top();
            v[i] = x;
            s.pop();
        }
    }
    for(auto i : v)
        cout << i;
    cout << endl;
    return 0;
}