Codeforces Round #672 (Div. 2)

A题

 

 容易题。只有严格单调递减情况是NO因为需要n*(n-1)/2次。其余都行

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#include <bits/stdc++.h> using namespace std; int main () {     int t;     cin >> t;     while(t--) {         int n;         cin >> n;         vector<int> v;         bool ok = 0;         for(int i = 0; i < n; ++i) {             int temp;             cin >> temp;             v.push_back(temp);         }         for(int i = 1; i < n; ++i) {             if(v[i - 1] <= v[i]) {                 ok = 1;             }         }         if(ok) {             cout << "YES" << endl;         }         else {             cout << "NO" << endl;         }     } }

 B题

记得开long long!!!!!!!!!!!!!!!!!!!!!!!!!!

 

 

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#include <bits/stdc++.h> using namespace std; typedef long long ll; int main () {     int t;     cin >> t;     while(t--) {         int n;         cin >> n;         vector<ll> v(100);         for(int i = 0; i < n; ++i) {             ll temp;             cin >> temp;             int circle = 0;             for(circle = 0; temp; temp >>= 1) {                 circle++;             }             v[circle]++;         }         ll ans = 0;         for(int i = 0; i < v.size(); ++i) {             if(v[i]) {                 ans +=  (v[i] - 1) * v[i] / 2;             }         }         cout << ans << endl;     } }

 C题

 

 思维题

用两个数组进行记录

一定要注意这种思想

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#include <bits/stdc++.h> using namespace std; typedef long long ll; int main () {     ios::sync_with_stdio(false);     cin.tie(0);     int t;     cin >> t;     while(t--) {         int n, q;         cin >> n >> q;         vector<ll> a(n + 1);         for(int i = 1; i <= n; ++i) {             cin >> a[i];         }         vector<ll> mx(n + 1);         vector<ll> mn(n + 1); //      for(int i = n - 1; i >= 0; --i) { //          mx[i] = max(mx[i + 1], a[i] - mn[i + 1]); //          mn[i] = min(mn[i + 1], a[i] - mx[i + 1]); //      } //      cout << mx[0] << endl;         for(int i = 1; i <= n; ++i) {             mx[i] = max(mx[i - 1], a[i] - mn[i - 1]);             mn[i] = min(mn[i - 1], a[i] - mx[i - 1]);         }         cout << mx[n] << endl;     } }

 D题

组合数问题 用优先队列解决。本质上来讲是求几个区间公共情况的组合情况

 

 

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#include <bits/stdc++.h> using namespace std; typedef long long ll; const int MODE = 998244353; const int N = 3e5 + 10; typedef pair<int, int> P; ll pre[N]; ll qpow(ll a, ll b) {     ll res = 1;     while(b) {         if(b & 1)             res = (res * a) % MODE;         b >>= 1;         a = (a * a) % MODE;     }     return res % MODE; } ll C(ll n, ll k) {     if(k == 0) {         return 1;     }     return(pre[n] % MODE * qpow((pre[k] * pre[n - k]) % MODE, MODE - 2) % MODE) % MODE; } int main () {     ll n, k;     cin >> n >> k;     pre[0] = 1;     for(int i = 1; i < N; ++i) {         pre[i] = (pre[i - 1] * i) % MODE;     }     vector<P> v(n);     for(int i = 0; i < n; ++i) {         cin >> v[i].first >> v[i].second;     }     sort(v.begin(), v.end()); //  for(int i = 0; i < v.size(); ++i) { //      cout << v[i].first << " " << v[i].second << endl; //  }     ll ans = 0; //  priority_queue<int, vector<int>, greater<int>> pq;     priority_queue<int, vector<int>, greater<int> > pq;     for(int i = 0; i < n; ++i) {         while(pq.size() && v[i].first > pq.top()) {             pq.pop();         }         if(pq.size() >= k - 1) {             ans += C(pq.size(), k - 1);         }         pq.push(v[i].second);     }     cout << ans % MODE << endl; }