Submission #920272

#TimeUsernameProblemLanguageResultExecution timeMemory
920272vjudge1Diversity (CEOI21_diversity)C++17
0 / 100
0 ms348 KiB
#include <iostream> #include <complex> #include <vector> #include <string> #include <algorithm> #include <cstdio> #include <numeric> #include <cstring> #include <ctime> #include <cstdlib> #include <set> #include <map> #include <unordered_map> #include <unordered_set> #include <list> #include <cmath> #include <bitset> #include <cassert> #include <queue> #include <stack> #include <deque> #include <random> using namespace std; template<typename T1, typename T2> inline void chkmin(T1 &a, T2 b) {if (a > b) a = b;} template<typename T1, typename T2> inline void chkmax(T1 &a, T2 b) {if (a < b) a = b;} #define files(FILENAME) read(FILENAME); write(FILENAME) #define read(FILENAME) freopen((FILENAME + ".in").c_str(), "r", stdin) #define write(FILENAME) freopen((FILENAME + ".out").c_str(), "w", stdout) #define all(c) (c).begin(), (c).end() #define sz(c) (int)(c).size() #define left left228 #define right right228 #define y1 y1228 #define mp make_pair #define pb push_back #define y2 y2228 #define rank rank228 using ll = long long; using ld = long double; const string FILENAME = "input"; const int MAXN = 300228; int n, q; int cnt[MAXN], f[MAXN]; ll func(int n) { return n * (n + 1) / 2; } ll calc(vector<int> a) { ll res = 0; ll x = 0; for (auto i: a) { res += func(x) + func(n - x - i); x += i; } res = func(n) * sz(a) - res; return res; } int main() { ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0); //read(FILENAME); cin >> n >> q; vector<int> a(n + 1); int mx = 0; for (int i = 1; i <= n; i++) { cin >> a[i]; chkmax(mx, a[i]); } for (int i = 1; i <= q; i++) { int l, r; cin >> l >> r; for (int j = l; j <= r; j++) { f[a[j]]++; } vector<int> v; for (int i = 1; i <= mx; i++) { if (f[i] != 0) { v.pb(f[i]); } } sort(all(v)); vector<int> optim; for (int i = 0; i < sz(v); i += 2) { optim.pb(v[i]); } for (int i = 1; i < sz(v); i += 2) { optim.pb(v[i]); } cout << calc(optim) << '\n'; for (int j = l; j <= r; j++) { f[a[j]] = 0; } } return 0; }
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