/ SeriousOJ /

Record Detail

Wrong Answer


  
# Status Time Cost Memory Cost
#1 Accepted 1ms 532.0 KiB
#2 Wrong Answer 6ms 496.0 KiB
#3 Accepted 2ms 576.0 KiB
#4 Wrong Answer 3ms 532.0 KiB

Code

//#pragma GCC target("avx2")
//#pragma GCC optimize("O3")
//#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int,int>;
using pll = pair<ll,ll>;
using pli = pair<ll,int>;
#define MOD 998244353
//#define MOD 1000000007
#define el '\n'
#define El '\n'
#define YESNO(x) ((x) ? "Yes" : "No")
#define YES YESNO(true)
#define NO YESNO(false)
#define EXIT_ANS(x) {cout << (x) << '\n'; return;}
#define PA() {EXIT_ANS(ans);}
template <typename T> void inline SORT(vector<T> &v){sort(v.begin(),v.end()); return;}
template <typename T> void inline REV(vector<T> &v){reverse(v.begin(),v.end()); return;}
template <typename T> void inline VEC_UNIQ(vector<T> &v){sort(v.begin(),v.end()); v.erase(unique(v.begin(),v.end()),v.end()); return;}
template <typename T> T inline MAX(vector<T> &v){return *max_element(v.begin(),v.end());}
template <typename T> T inline MIN(vector<T> &v){return *min_element(v.begin(),v.end());}
template <typename T> T inline SUM(vector<T> &v){T ans = 0; for(int i = 0; i < (int)v.size(); i++)ans += v[i]; return ans;}
template <typename T> void inline DEC(vector<T> &v){for(int i = 0; i < (int)v.size(); i++)v[i]--; return;}
template <typename T> void inline INC(vector<T> &v){for(int i = 0; i < (int)v.size(); i++)v[i]++; return;}
void inline TEST(void){cerr << "TEST" << endl; return;}
template <typename T> bool inline chmin(T &x,T y){
    if(x > y){
        x = y;
        return true;
    }
    return false;
}
template <typename T> bool inline chmax(T &x,T y){
    if(x < y){
        x = y;
        return true;
    }
    return false;
}
template <typename T = long long> vector<T> inline get_vec(int n){
    vector<T> ans(n);
    for(int i = 0; i < n; i++)cin >> ans[i];
    return ans;
}
template <typename T> void inline print_vec(vector<T> &vec,bool kaigyou = false){
    int n = (int)vec.size();
    for(int i = 0; i < n; i++){
        cout << vec[i];
        if(kaigyou || i == n - 1)cout << '\n';
        else cout << ' ';
    }
    if(!n)cout << '\n';
    return;
}
template <typename T> void inline debug_vec(vector<T> &vec,bool kaigyou = false){
    int n = (int)vec.size();
    for(int i = 0; i < n; i++){
        cerr << vec[i];
        if(kaigyou || i == n - 1)cerr << '\n';
        else cerr << ' ';
    }
    if(!n)cerr << '\n';
    return;
}
vector<vector<int>> inline get_graph(int n,int m = -1,bool direct = false){
    if(m == -1)m = n - 1;
    vector<vector<int>> g(n);
    while(m--){
        int u,v;
        cin >> u >> v;
        u--; v--;
        g[u].push_back(v);
        if(!direct)g[v].push_back(u);
    }
    return g;
}
template <typename T> vector<vector<pair<T,int>>> inline get_weighted_graph(int n,int m = -1,bool direct = false){
    if(m == -1)m = n - 1;
    vector<vector<pair<T,int>>> g(n);
    while(m--){
        int u,v;
        cin >> u >> v;
        u--; v--;
        ll w; cin >> w;
        g[u].push_back(pair(w,v));
        if(!direct)g[v].push_back(pair(w,u));
    }
    return g;
}

// 抽象再帰セグ木?(抽象セグ木の定義分からんな)(遅延ではない)
template <typename T> class ococo_segtree {
    T e = INT_MIN / 2;
    T func(T a, T b) {
        // ここに演算を入れる
        return max(a,b);
    }

    T get_minimum2(int a, int b, int k, int l, int r) {
        if(a <= l && r <= b) return rmq[k];
        else if(r <= a || b <= l) return tmax;
        else return func(get_minimum2(a, b, k * 2 + 1, l, (l + r) / 2), get_minimum2(a, b, k * 2 + 2, (l + r) / 2, r));
    }

  public:
    int n;
    T tmax;
    // range minimum query
    vector<T> rmq;

    //(データの大きさ,単位元)
    ococo_segtree(int N) {
        syokica(N);
    }

    // 配列の初期化 O(N)
    void syokica(int a) {
        n = 1;
        tmax = e;
        while(n < a) n *= 2;
        rmq.resize(2 * n - 1);
        for(int i = 0; i < 2 * n - 1; i++) {
            rmq[i] = tmax;
        }
    }

    // a[i]をxにする O(logN)
    void update_num(int i, T x) {
        i += (n - 1);
        rmq[i] = x;
        while(i) {
            i = (i - 1) / 2;
            rmq[i] = func(rmq[i * 2 + 1], rmq[i * 2 + 2]);
        }
    }

    // a[i]にxを加える O(logN)
    void add_num(int i, T x) {
        i += (n - 1);
        rmq[i] += x;
        while(i) {
            i = (i - 1) / 2;
            rmq[i] = func(rmq[i * 2 + 1], rmq[i * 2 + 2]);
        }
    }

    
    //[l,r]の区間演算の取得 O(logN)
    T get_val(int l, int r) {
        return get_minimum2(l, r + 1, 0, 0, n);
    }
};


#define MULTI_TEST_CASE true
void solve(void){
    //問題を見たらまず「この問題設定から言えること」をいっぱい言う
    //よりシンプルな問題に言い換えられたら、言い換えた先の問題を自然言語ではっきりと書く
    //複数の解法のアイデアを思いついた時は全部メモしておく
    //g++ -D_GLIBCXX_DEBUG -Wall -O2 f.cpp -o o
    int n;
    cin >> n;
    vector<ll> a = get_vec(n);
    vector<ll> c = get_vec(n);
    if(n == 1)EXIT_ANS(0);

    //p[i] = a[0]+a[1]+a[2] + ... + a[i]

    //l -> r
    //p[r] - p[l-1] >= c[r]
    //p[r]-c[r] >= p[l-1]

    vector<ll> p(n);
    p[0] = a[0];
    for(int i = 1; i < n; i++){
        p[i] = p[i - 1] + a[i];
    }
    
    vector<ll> pc(n);
    for(int i = 0; i < n; i++)pc[i] = p[i] - c[i];
    vector<ll> vec = pc;
    vec.push_back(0LL);
    for(int i = 0; i < n; i++)vec.push_back(p[i]);

    // debug_vec(p);
    // debug_vec(pc);

    VEC_UNIQ(vec);
    int m = (int)vec.size();
    
    //dp[i] = 点 i に行くとして、何回の移動で行けるか
    vector<int> dp(n,INT_MIN / 2);

    ococo_segtree<int> sgt(m + 1);

    int tempidx = lower_bound(vec.begin(),vec.end(),0) - vec.begin();
    sgt.update_num(tempidx,dp[0] = 0);
    
    for(int i = 1; i < n; i++){
        int pidx = lower_bound(vec.begin(),vec.end(),p[i - 1]) - vec.begin();
        int pcidx = lower_bound(vec.begin(),vec.end(),pc[i]) - vec.begin();
        int temp = sgt.get_val(0,pcidx);

        dp[i] = temp + 1;
        sgt.update_num(pidx,dp[i]);
    }
    //debug_vec(dp);
    int ans = max(dp[n - 1],-1);
    cout << ans << el;

    return;
}

void calc(void){
    return;
}

signed main(void){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    calc();
    int t = 1;
    if(MULTI_TEST_CASE)cin >> t;
    while(t--){
        solve();
    }
    return 0;
}

Information

Submit By
Type
Submission
Problem
P1199 F. Roy and Path Game
Language
C++17 (G++ 13.2.0)
Submit At
2026-01-08 18:17:37
Judged At
2026-01-08 18:17:37
Judged By
Score
5
Total Time
6ms
Peak Memory
576.0 KiB