結果
問題 | No.1705 Mode of long array |
ユーザー |
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提出日時 | 2022-12-09 01:27:26 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 348 ms / 3,000 ms |
コード長 | 3,607 bytes |
コンパイル時間 | 1,892 ms |
コンパイル使用メモリ | 185,464 KB |
実行使用メモリ | 23,552 KB |
最終ジャッジ日時 | 2024-10-14 18:27:00 |
合計ジャッジ時間 | 15,564 ms |
ジャッジサーバーID (参考情報) |
judge4 / judge3 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 51 |
ソースコード
#pragma GCC optimization ("O3")#include <bits/stdc++.h>using namespace std;using ll = long long;using vec = vector<ll>;using mat = vector<vec>;using pll = pair<ll,ll>;#define INF (1LL<<61)#define MOD 1000000007LL// #define MOD 998244353LL#define EPS (1e-10)#define PR(x) cout << (x) << endl#define PS(x) cout << (x) << " "#define REP(i,m,n) for(ll (i)=(m),(i_len)=(n);(i)<(i_len);++(i))#define FORE(i,v) for(auto (i):v)#define ALL(x) (x).begin(), (x).end()#define SZ(x) ((ll)(x).size())#define REV(x) reverse(ALL((x)))#define ASC(x) sort(ALL((x)))#define DESC(x) {ASC((x)); REV((x));}#define BIT(s,i) (((s)>>(i))&1)#define pb push_back#define fi first#define se secondtemplate<class T> inline int chmin(T& a, T b) {if(a>b) {a=b; return 1;} return 0;}template<class T> inline int chmax(T& a, T b) {if(a<b) {a=b; return 1;} return 0;}class mint {public:ll x;mint(ll x=0) : x((x%MOD+MOD)%MOD) {}mint operator-() const {return mint(-x);}mint& operator+=(const mint& a) {if((x+=a.x)>=MOD) x-=MOD; return *this;}mint& operator-=(const mint& a) {if((x+=MOD-a.x)>=MOD) x-=MOD; return *this;}mint& operator*=(const mint& a) {(x*=a.x)%=MOD; return *this;}mint operator+(const mint& a) const {mint b(*this); return b+=a;}mint operator-(const mint& a) const {mint b(*this); return b-=a;}mint operator*(const mint& a) const {mint b(*this); return b*=a;}mint pow(ll t) const {if(!t) return 1; mint a=pow(t>>1); return (t&1?*this*a:a)*a;}mint inv() const {return pow(MOD-2);}mint& operator/=(const mint& a) {return *this*=a.inv();}mint operator/(const mint& a) const {mint b(*this); return b/=a;}};istream &operator>>(istream& is, mint& a) {ll t; is>>t; a=t; return is;}ostream &operator<<(ostream& os, const mint& a) {return os<<a.x;}template <typename X> struct SegmentTree {using FX = function<X(X, X)>;int n;FX fx;const X ex;vector<X> dat;SegmentTree(int n_, FX fx_, X ex_): n(), fx(fx_), ex(ex_) {int x = 1;while(n_ > x) x *= 2;n = x;dat = vector<X>(n*2, ex);}void set(int i, X x) {dat[i+n-1] = x;}void build() {for (int k = n-2; k>=0; k--) dat[k] = fx(dat[k*2+1], dat[k*2+2]);}void update(int i, X x) {i += n-1;dat[i] = x;while(i > 0) {i = (i-1)/2;dat[i] = fx(dat[i*2+1], dat[i*2+2]);}}X output_sub(int a, int b, int k, int l, int r) {if(r <= a || b <= l) return ex;else if(a <= l && r <= b) return dat[k];else {X vl = output_sub(a, b, k*2+1, l, (l+r)/2);X vr = output_sub(a, b, k*2+2, (l+r)/2, r);return fx(vl, vr);}}X output(int a, int b) {return output_sub(a, b, 0, 0, n);}};int main(){ll N, M;cin >> N >> M;ll ex = -INF;auto fx = [](ll l, ll r) -> ll {return max(l, r);};SegmentTree<ll> st(M+1, fx, ex);map<ll,set<ll>> S;REP(i,1,M+1) {ll a;cin >> a;st.set(i, a);S[a].insert(i);}st.build();ll Q;cin >> Q;while(Q--) {ll t, x, y;cin >> t >> x >> y;if(t <= 2) {ll a = st.output(x, x+1);ll b = a+y*(t==1?1:-1);S[a].erase(x);S[b].insert(x);st.update(x, b);}else {ll m = st.output(1, M+1);PR(*(S[m].rbegin()));}}return 0;}/**/