Codeforces Round 870 (Div. 2)题解（5/6)

于 2023-05-06 13:14:14 发布
阅读量608 收藏
点赞数
分类专栏： Codeforces 比赛题解 文章标签： c++ 算法 图论 codeforces
版权声明：本文为博主原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接和本声明。
本文链接：https://blog.youkuaiyun.com/weixin_45539557/article/details/130524858
版权
文章包含多个编程竞赛题目，涉及逻辑推理和算法设计。A题通过暴力枚举判断说谎人数；B题利用回文性质求解最大x值；C题探讨投票平局条件；D题计算区间最大值问题；E题在约束下求费用最大模型。每道题都提出了相应的解题思路和解决方案。

摘要生成于 C知道 ，由 DeepSeek-R1 满血版支持， 前往体验 >

A.Trust Nobody

题意：
n n n个人，有人说谎，有人说真话，每个人都告诉你至少有 l i l_i li​个人说谎，问是否矛盾，否则输出可能的说谎的人数。

思路：
暴力枚举说谎人数，判断有多少个人说谎即可。

#include <bits/stdc++.h>
#define local
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
using ll = long long;
using db = double;
using PII = pair<int, int>;
using PLI = pair<ll, int>;

template<class T>
ostream &operator<<(ostream &io, vector<T> a) {
    io << "{"; 
    for (auto I: a)io << I << " ";
    io << "}";
    return io;
}

template<class T>
ostream &operator<<(ostream &io, set<T> a) {
    io << "{"; 
    for (auto I: a)io << I << " ";
    io << "}";
    return io;
}

template<class K, class V>
ostream &operator<<(ostream &io, map<K, V> a) {
    io << "(";
    for (auto I: a)io << "{" << I.first << ":" << I.second << "}";
    io << ")"; 
    return io;
}

void debug_out() {
    cerr << endl;
}

template<typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
    cerr << ' ' << H;
    debug_out(T...);
}

#ifdef local
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 1
#endif
#define all(x) x.begin(),x.end()


int main() {
    IOS
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        vector<int> a(n + 1);
        for (int i = 1; i <= n; ++i) {
            int x;
            cin >> x;
            a[x]++;
        }
        for (int i = 1; i <= n; ++i) {
            a[i] +=  a[i - 1];
        }
        bool f = 0;
        for (int i = 0; i <= n; ++i) {
            if (i == n - a[i]) {
                f = 1;
                cout << i << "\n";
                break;
            }
        }
        if (!f) {
            cout << -1 << '\n';
        }
    }
    return 0;
}

B.Lunatic Never Content

题意：
f ( a , x ) = ( a 1 m o d x . . . . a n m o d x ) f(a, x) = (a1 mod x....anmodx) f(a,x)=(a1modx....anmodx)，问最大的x使得成回文。

思路：考虑 a m o d x = b m o d x a mod x = b mod x amodx=bmodx， 且 x x x属于 [ a , b ] [a,b] [a,b],不难推出 t x = a b s ( a − b ) tx = abs(a -b) tx=abs(a−b), 所以每个限制等于告诉你我都要是 a b s ( a − b ) abs(a - b) abs(a−b)的约数，做 g c d gcd gcd即可。

#include <bits/stdc++.h>
#define local
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
using ll = long long;
using db = double;
using PII = pair<int, int>;
using PLI = pair<ll, int>;

template<class T>
ostream &operator<<(ostream &io, vector<T> a) {
    io << "{"; 
    for (auto I: a)io << I << " ";
    io << "}";
    return io;
}

template<class T>
ostream &operator<<(ostream &io, set<T> a) {
    io << "{"; 
    for (auto I: a)io << I << " ";
    io << "}";
    return io;
}

template<class K, class V>
ostream &operator<<(ostream &io, map<K, V> a) {
    io << "(";
    for (auto I: a)io << "{" << I.first << ":" << I.second << "}";
    io << ")"; 
    return io;
}

void debug_out() {
    cerr << endl;
}

template<typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
    cerr << ' ' << H;
    debug_out(T...);
}

#ifdef local
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 1
#endif
#define all(x) x.begin(),x.end()


int main() {
    IOS
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        vector<int> a(n + 2);
        for (int i = 1; i <= n; ++i) {
            cin >> a[i];
        }
        int ans = 0;
        for (int i = 1, j = n; i <= j; i++, j--) {
            ans = __gcd(ans, abs(a[i] - a[j]));
        }
        cout << ans << '\n';
    }
    return 0;
}

C.Dreaming of Freedom

题意： n n n个人投票，给 m m m个人，问是否存在永远平局的情况。

思路：
n < = m n <= m n<=m, 显然可以，除了 n = = 1 n == 1 n==1,
否则枚举 n n n的约数a看 m > = a m >= a m>=a是否成立即可

#include <bits/stdc++.h>
#define local
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
using ll = long long;
using db = double;
using PII = pair<int, int>;
using PLI = pair<ll, int>;

template<class T>
ostream &operator<<(ostream &io, vector<T> a) {
    io << "{"; 
    for (auto I: a)io << I << " ";
    io << "}";
    return io;
}

template<class T>
ostream &operator<<(ostream &io, set<T> a) {
    io << "{"; 
    for (auto I: a)io << I << " ";
    io << "}";
    return io;
}

template<class K, class V>
ostream &operator<<(ostream &io, map<K, V> a) {
    io << "(";
    for (auto I: a)io << "{" << I.first << ":" << I.second << "}";
    io << ")"; 
    return io;
}

void debug_out() {
    cerr << endl;
}

template<typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
    cerr << ' ' << H;
    debug_out(T...);
}

#ifdef local
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 1
#endif
#define all(x) x.begin(),x.end()


int main() {
    IOS
    int t;
    cin >> t;
    while (t--) {
        int n, m;
        cin >> n >> m;
        if (n == 1) {
            cout << "YES\n";
        } else if (n <= m) {
            cout << "NO\n";
        } else {
            bool f = 0;
            for (int i = 2; i * i <= n; ++i) {
                if (n % i == 0) {
                    if (m >= i) {
                       f = 1;
                       break; 
                    }
                }
            }
            cout << (f ? "NO\n" : "YES\n");
        }
    }
    return 0;
}

D.Dreaming of Freedom

题意:
定义 f ( l , r ) f(l, r) f(l,r) = (区间最大的三个值 - ( r − l ) ) (r - l)) (r−l))， 问最大的 f ( l , r ) f(l, r) f(l,r)
思路：

  • 显然区间的左右端点必选，所以左右端点贡献分别为 a [ i ] + i a[i]+i a[i]+i, a [ i ] − i a[i]-i a[i]−i,暴力枚举中间端点分别对这两个贡献数组做前后缀最大值即可。
  • 考虑类似银川 B B B的 t r i c k trick trick, 由于值的数量很少，我们暴力dp出其贡献，具体来说就是 d p [ i ] [ 0 / 1 / 2 / 3 ] dp[i][0/1/2/3] dp[i][0/1/2/3], 表示当前区间选了前三个数/or不选的最大值，左右端点的贡献为 a [ i ] + / − i a[i] +/- i a[i]+/−i, 暴力转移即可。
#include <bits/stdc++.h>
#define local
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
using ll = long long;
using db = double;
using PII = pair<int, int>;
using PLI = pair<ll, int>;

template<class T>
ostream &operator<<(ostream &io, vector<T> a) {
    io << "{"; 
    for (auto I: a)io << I << " ";
    io << "}";
    return io;
}

template<class T>
ostream &operator<<(ostream &io, set<T> a) {
    io << "{"; 
    for (auto I: a)io << I << " ";
    io << "}";
    return io;
}

template<class K, class V>
ostream &operator<<(ostream &io, map<K, V> a) {
    io << "(";
    for (auto I: a)io << "{" << I.first << ":" << I.second << "}";
    io << ")"; 
    return io;
}

void debug_out() {
    cerr << endl;
}

template<typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
    cerr << ' ' << H;
    debug_out(T...);
}

#ifdef local
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 1
#endif
#define all(x) x.begin(),x.end()


int main() {
    IOS
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        vector<int> b(n + 2), sum1(n + 2), sum2(n + 2);
        for (int i = 1; i <= n; ++i) {
            cin >> b[i];
            sum1[i] = b[i] + i;
            sum2[i] = b[i] - i;
        }
        for (int i = 2; i <= n; ++i) {
            sum1[i] = max(sum1[i - 1], sum1[i]);
        }
        for (int i = n - 1; i >= 1; --i) {
            sum2[i] = max(sum2[i + 1], sum2[i]);
        }
        int ans = 0;
        for (int i = 2; i + 1 <= n; ++i) {
            ans = max(ans, b[i] + sum1[i - 1] + sum2[i + 1]);
        }
        cout << ans << '\n';
    }
    return 0;
}

#include <bits/stdc++.h>
#define local
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
using ll = long long;
using db = double;
using PII = pair<int, int>;
using PLI = pair<ll, int>;

template<class T>
ostream &operator<<(ostream &io, vector<T> a) {
    io << "{"; 
    for (auto I: a)io << I << " ";
    io << "}";
    return io;
}

template<class T>
ostream &operator<<(ostream &io, set<T> a) {
    io << "{"; 
    for (auto I: a)io << I << " ";
    io << "}";
    return io;
}

template<class K, class V>
ostream &operator<<(ostream &io, map<K, V> a) {
    io << "(";
    for (auto I: a)io << "{" << I.first << ":" << I.second << "}";
    io << ")"; 
    return io;
}

void debug_out() {
    cerr << endl;
}

template<typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
    cerr << ' ' << H;
    debug_out(T...);
}

#ifdef local
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 1
#endif
#define all(x) x.begin(),x.end()

const int maxn = 1e5 + 5;
int dp[maxn][4];

int main() {
    IOS
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        for (int i = 1, x; i <= n; ++i) {
            cin >> x;
            for (int j = 0; j <= 3; ++j)
                dp[i][j] = dp[i - 1][j];
            dp[i][1] = max(dp[i][1], dp[i - 1][0] + x + i);
            dp[i][2] = max(dp[i][2], dp[i - 1][1] + x);
            dp[i][3] = max(dp[i][3], dp[i - 1][2] + x - i);
        }
        cout << dp[n][3] << '\n';
    }
    return 0;
}

E.Dreaming of Freedom

题意: m m m场演出， n n n个模特，每个模特一个费用， m m m场演出每个模特有不同的rank, 要求选出模特的子集后重排列使得 m m m场演出的rank严格递增，且费用最大。

思路：
显然有个 n 2 n^2 n2的 d p dp dp, d p [ i ] dp[i] dp[i]表示最后结尾的是 i i i这个模特的最大值，如果知道每个模特后面能跟谁，就相当于在一个dag上 d p dp dp, O ( n 2 ) O(n^2) O(n2)即可，这个每个模特后面能跟谁也显然有个 O ( n 2 m ) O(n^2 m) O(n2m)的做法，两两枚举 n n n，但是这样无法优化。我们发现这个关系 a < b ， b < c 则 a < c a < b，b < c则 a < c a<b，b<c则a<c, 满足偏序关系，所以排序后一定是dag图，那么我们考虑对每一个 m m m做倒序排序， f [ i ] [ j ] f[i][j] f[i][j]表示 i i i后面能跟谁，我们在这个 d a g dag dag图上转移是 O ( n ) O(n) O(n)的，我们可以通过dag图上转移使得一次转移第二维的所有位数，第二维 b i t s e t bitset bitset优化即可做到O(n/64), 最后按拓扑序再 d p dp dp一下即可。
时间复杂度 O ( n 2 m / 64 ) O(n ^ 2m/64) O(n2m/64)，好久没 b i t s e t bitset bitsetvp的时候差点没写完，该找时间写些 b i t s e t bitset bitset的题了

#include <bits/stdc++.h>
#define local
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
using ll = long long;
using db = double;
using PII = pair<int, int>;
using PLI = pair<ll, int>;

template<class T>
ostream &operator<<(ostream &io, vector<T> a) {
    io << "{"; 
    for (auto I: a)io << I << " ";
    io << "}";
    return io;
}

template<class T>
ostream &operator<<(ostream &io, set<T> a) {
    io << "{"; 
    for (auto I: a)io << I << " ";
    io << "}";
    return io;
}

template<class K, class V>
ostream &operator<<(ostream &io, map<K, V> a) {
    io << "(";
    for (auto I: a)io << "{" << I.first << ":" << I.second << "}";
    io << ")"; 
    return io;
}

void debug_out() {
    cerr << endl;
}

template<typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
    cerr << ' ' << H;
    debug_out(T...);
}

#ifdef local
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 1
#endif
#define all(x) x.begin(),x.end()

const int maxn = 5005;
bitset<5000> dp[maxn];
PII a[maxn];

vector<int> G[maxn];
int deg[maxn];

int main() {
    IOS
    int m, n;
    cin >> m >> n;
    vector<int> p(n + 1);
    for (int i = 1; i <= n; ++i) {
        cin >> p[i];
    }
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
            if (i != j) {
                dp[i].set(j);
            }
        }
    }
    for (int i = 1; i <= m; ++i) {
        for (int j = 1; j <= n; ++j) {
            a[j].second = j - 1;
            cin >> a[j].first;
        }
        sort(a + 1, a + 1 + n, [&](PII x, PII y) {
            return x.first > y.first;
        });
        bitset<5000> b;
        vector<int> tmp;
        for (int j = 1; j <= n; ++j) {
            tmp.emplace_back(a[j].second);
            if (j == n || a[j].first != a[j + 1].first) {
                for (auto &x : tmp) {
                    dp[x] &= b;
                }
                for (auto &x : tmp) {
                    b.set(x);
                }
                tmp.clear();
            }
        }
    }
    for (int i = 0; i < n; ++i) {
        for (int j = dp[i]._Find_first(); j < n; j = dp[i]._Find_next(j)) { 
                G[i].emplace_back(j);
                deg[j]++;
        }
    }
    queue<int> q;
    vector<ll> f(n + 2, 0);
    for (int i = 0; i < n; ++i) {
        if (!deg[i]) {
            f[i] = p[i  + 1];
            q.push(i);
        }
    }
    while (q.size()) {
        int x = q.front();
        q.pop();
        for (auto u : G[x]) {
            f[u] = max(f[u], f[x] + p[u + 1]);
            if (--deg[u] == 0) {
                q.push(u);
            }
        }
    }
    ll ans = 0;
    for (int i = 0; i < n; ++i) {
        ans = max(ans, f[i]);
    }
    cout << ans;
    return 0;
}

F.Fading into Fog

待补

评论 1
添加红包

请填写红包祝福语或标题

个

红包个数最小为10个

元

红包金额最低5元

当前余额3.43元 前往充值 >
需支付：10.00元
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付元
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明：

1.余额是钱包充值的虚拟货币，按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载，可以购买VIP、付费专栏及课程。

余额充值