답안 #751499

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
751499 2023-05-31T16:28:12 Z JakobZorz Sure Bet (CEOI17_sure) C++14
60 / 100
2000 ms 3160 KB
#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
#include <stack>
#include <limits.h>
#include <math.h>
#include <iomanip>
#include <bitset>
#include <unordered_map>
#include <map>
 
//#pragma GCC target("popcnt")

#define ll long long
using namespace std;
const int MOD = 1e9 + 7;

//#define SEGTREE
//#define TREE
//#define DSU
#define MATH
 
#ifdef SEGTREE
template<class Type>
class SegmentTree {
    Type (*func)(Type a, Type b);
    vector<Type> nodes;
    vector<int> l;
    vector<int> r;
    int size_log2;
    Type neutral;
    
    void init_node(int node) {
        if(node >= (1 << size_log2))
            return;
        
        l[2 * node] = l[node];
        r[2 * node] = (l[node] + r[node]) / 2;
        init_node(2 * node);
        
        l[2 * node + 1] = (l[node] + r[node]) / 2;
        r[2 * node + 1] = r[node];
        init_node(2 * node + 1);
        
        nodes[node] = func(nodes[2 * node], nodes[2 * node + 1]);
    }
    
    void update_node(int node) {
        if(node < (1 << size_log2))
            nodes[node] = func(nodes[2 * node], nodes[2 * node + 1]);
        
        if(node != 1)
            update_node(node / 2);
    }
    
    Type get(int node, int ll_, int rr) {
        if(rr <= l[node] || ll_ >= r[node])
            return neutral;
        
        if(ll_ <= l[node] && rr >= r[node])
            return nodes[node];
        
        Type left = get(2 * node, ll_, rr);
        Type right = get(2 * node + 1, ll_, rr);
        
        return func(left, right);
    }
    
public:
    SegmentTree(Type (*func)(Type a, Type b), vector<Type> vector, Type neutral) : func(func), neutral(neutral) {
        size_log2 = 0;
        while((1 << size_log2) < vector.size())
            size_log2++;
        
        nodes.resize((1 << size_log2) * 2);
        l.resize((1 << size_log2) * 2);
        r.resize((1 << size_log2) * 2);
        
        for(int i = 0; i < vector.size(); i++)
            nodes[(1 << size_log2) + i] = vector[i];
        
        l[1] = 0;
        r[1] = 1 << size_log2;
        init_node(1);
    }
    
    void set(int idx, Type el) {
        nodes[(1 << size_log2) + idx] = el;
        update_node((1 << size_log2) + idx);
    }
    
    Type get(int l, int r) {
        return get(1, l, r);
    }
    
    Type get(int idx) {
        return nodes[(1 << size_log2) + idx];
    }
    
    Type get() {
        return nodes[1];
    }
    
    int get_first_larger_or_eq_than(int bound) {
        return get_first_larger_or_eq_than(1, bound);
    }
    
    int get_first_larger_or_eq_than(int node, int bound) {
        if(node >= (1 << size_log2)) {
            return node - (1 << size_log2);
        }
        
        if(nodes[node * 2] > bound)
            return get_first_larger_or_eq_than(node * 2, bound);
        else
            return get_first_larger_or_eq_than(node * 2 + 1, bound - nodes[node * 2]);
    }
};
 
#endif
 
#ifdef TREE
#define POW 18
 
struct Node {
    int parents[POW];
    vector<int> conns;
    int depth;
    int value;
    int subtree_depth;
    int subtree_size;
    int euler;
    int val;
    int res;
};

ll add(ll a, ll b) {
    return a + b;
}

Node nodes[200000];
int n;
 
int setup(int node, int depth, int euler, ll dist) {
    dist += nodes[node].value;
    nodes[node].depth = depth;
    nodes[node].euler = euler++;
    
    
    for(int i = 1; i < POW; i++) {
        nodes[node].parents[i] = nodes[nodes[node].parents[i - 1]].parents[i - 1];
    }
    
    int size = 1;
    
    for(int i = 0; i < nodes[node].conns.size(); i++) {
        int child = nodes[node].conns[i];
        if(child != nodes[node].parents[0]) {
            nodes[child].parents[0] = node;
            euler = setup(child, depth + 1, euler, dist);
            size += nodes[child].subtree_size;
        }
    }
    nodes[node].subtree_size = size;
    return euler;
}
 
void setup_tree(int root) {
    nodes[root].parents[0] = root;
    setup(root, 0, 0, 0);
}
 
int get_subtree_depth(int node) {
    if(nodes[node].subtree_depth)
        return nodes[node].subtree_depth;
    
    int max_depth = 1;
    for(int child : nodes[node].conns) {
        if(child == nodes[node].parents[0])
            continue;
        max_depth = max(max_depth, get_subtree_depth(child) + 1);
    }
    nodes[node].subtree_depth = max_depth;
    return max_depth;
}
 
int get_parent(int node, int depth) {
    if(depth > nodes[node].depth)
        return -1;
    
    int climber = node;
    for(int i = 0; i < POW; i++) {
        if(depth & (1 << i) && climber != -1)
            climber = nodes[climber].parents[i];
    }
    
    return climber;
}
 
int lca(int a, int b) {
    if(nodes[a].depth < nodes[b].depth)
        swap(a, b);
    
    a = get_parent(a, nodes[a].depth - nodes[b].depth);
    
    if(a == b)
        return a;
    
    for(int i = POW - 1; i >= 0; i--) {
        if(nodes[a].parents[i] != nodes[b].parents[i]) {
            a = nodes[a].parents[i];
            b = nodes[b].parents[i];
        }
    }
    
    return nodes[a].parents[0];
}
 
#endif
 
#ifdef DSU
 
class Dsu {
    vector<int> arr;
    int num_sets;
    
public:
    Dsu(int size) {
        arr = vector<int>(size, -1);
        num_sets = size;
    }
    
    int merge(int a, int b) {
        a = get(a);
        b = get(b);
        
        if(a == b)
            return a;
        
        if(arr[a] > arr[b])
            swap(a, b);
        
        arr[a] += arr[b];
        arr[b] = a;
        num_sets--;
        return a;
    }
    
    int get(int a) {
        if(arr[a] < 0)
            return a;
        arr[a] = get(arr[a]);
        return get(arr[a]);
    }
    
    int get_size(int a) {
        return -arr[get(a)];
    }
    
    int get_num_sets() {
        return num_sets;
    }
};
 
#endif

#ifdef MATH
 
ll dpf[2000001];
ll factorial(ll n) {
    if(n == 0)
        return 1;
    
    if(dpf[n])
        return dpf[n];
    
    ll result = factorial(n - 1) * n;
    result %= MOD;
    dpf[n] = result;
    return result;
}
 
ll powm(ll base, ll exp) {
    ll result = 1;
    
    for(int i = 0; i < 64; i++) {
        if((exp >> i) % 2 == 1) {
            result *= base;
            result %= MOD;
        }
        base *= base;
        base %= MOD;
    }
    
    return result;
}
 
ll inverse(ll n) {
    return powm(n, MOD - 2);
}
 
ll dpif[2000001];
ll inverse_factorial(ll n) {
    if(dpif[n])
        return dpif[n];
    
    ll result = inverse(factorial(n));
    dpif[n] = result;
    return result;
}

ll choose(ll n, ll k) {
    return (((factorial(n)*inverse_factorial(n-k))%MOD)*inverse_factorial(k))%MOD;
}

ll gcd(ll a, ll b){
    if(a==b)
        return a;
    if(a<b)
        swap(a,b);
    if(b==0)
        return a;
    return gcd(b, a%b);
}

#endif

int main(){
    ios::sync_with_stdio(false);
    cout.tie(NULL);
    cin.tie(NULL);
     
    //freopen("mootube.in","r",stdin);
    //freopen("mootube.out","w",stdout);
    
    
    int n;
    cin>>n;
    vector<double>a1(n+1);
    vector<double>a2(n+1);
    for(int i=0;i<n;i++)
        cin>>a1[i]>>a2[i];
    sort(a1.begin(),a1.end());
    reverse(a1.begin(),a1.end());
    sort(a2.begin(),a2.end());
    reverse(a2.begin(),a2.end());
    
    for(int i=1;i<=n;i++){
        a1[i]+=a1[i-1];
        a2[i]+=a2[i-1];
    }
    
    double r=0;
    
    for(int i=0;i<=n;i++){
        for(int j=0;j<=n;j++){
            double c=min(a1[i],a2[j])-i-j-2;
            r=max(r,c);
        }
    }
    
    printf("%.4lf\n",(double)r);
    
    return 0;
}

/*
 
4
1.4 3.7
1.2 2
1.6 1.4
1.9 1.5
 
 
res:
 0.5000
 
 */
# 결과 실행 시간 메모리 Grader output
1 Correct 0 ms 212 KB Output is correct
2 Correct 1 ms 212 KB Output is correct
3 Correct 1 ms 212 KB Output is correct
4 Correct 0 ms 212 KB Output is correct
5 Correct 1 ms 212 KB Output is correct
6 Correct 1 ms 212 KB Output is correct
# 결과 실행 시간 메모리 Grader output
1 Correct 0 ms 212 KB Output is correct
2 Correct 1 ms 212 KB Output is correct
3 Correct 1 ms 212 KB Output is correct
4 Correct 0 ms 212 KB Output is correct
5 Correct 1 ms 212 KB Output is correct
6 Correct 1 ms 212 KB Output is correct
7 Correct 1 ms 212 KB Output is correct
8 Correct 1 ms 212 KB Output is correct
9 Correct 1 ms 212 KB Output is correct
10 Correct 1 ms 212 KB Output is correct
11 Correct 1 ms 212 KB Output is correct
12 Correct 4 ms 340 KB Output is correct
13 Correct 2 ms 340 KB Output is correct
14 Correct 2 ms 348 KB Output is correct
15 Correct 2 ms 348 KB Output is correct
16 Correct 2 ms 348 KB Output is correct
# 결과 실행 시간 메모리 Grader output
1 Correct 0 ms 212 KB Output is correct
2 Correct 1 ms 212 KB Output is correct
3 Correct 1 ms 212 KB Output is correct
4 Correct 0 ms 212 KB Output is correct
5 Correct 1 ms 212 KB Output is correct
6 Correct 1 ms 212 KB Output is correct
7 Correct 1 ms 212 KB Output is correct
8 Correct 1 ms 212 KB Output is correct
9 Correct 1 ms 212 KB Output is correct
10 Correct 1 ms 212 KB Output is correct
11 Correct 1 ms 212 KB Output is correct
12 Correct 4 ms 340 KB Output is correct
13 Correct 2 ms 340 KB Output is correct
14 Correct 2 ms 348 KB Output is correct
15 Correct 2 ms 348 KB Output is correct
16 Correct 2 ms 348 KB Output is correct
17 Execution timed out 2080 ms 3160 KB Time limit exceeded
18 Halted 0 ms 0 KB -