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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#136526#6401. Classic: N Real DNA Potsammardab3an#WA 1ms3856kbC++203.7kb2023-08-09 00:05:422023-08-09 00:05:46

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-08-09 00:05:46]
  • 评测
  • 测评结果:WA
  • 用时:1ms
  • 内存:3856kb
  • [2023-08-09 00:05:42]
  • 提交

answer


// By AmmarDab3an 

#include "bits/stdc++.h"

using namespace std;

#define int int64_t
#define ll  int64_t

// typedef unsigned int        uint;
// typedef long long int       ll;
// typedef unsigned long long  ull;
typedef pair<int, int>    pii;
typedef pair<ll, ll>      pll;
typedef pair<int, pii>    iii;
typedef pair<ll, pll>     lll;
typedef vector<int>       vi;
typedef vector<ll>        vl;
typedef vector<pii>       vpii;
typedef vector<pll>       vpll;

#define endl '\n'
#define fastIO ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define freopenI freopen("input.txt", "r", stdin);
#define freopenO freopen("output.txt", "w", stdout);

const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-9;
const double  PI = acos(-1);

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
 
int rand(int x, int y) {
	return uniform_int_distribution<int>(x, y)(rng);
}

int mul(int a, int b){
	int ret = (1ll * (a%MOD) * (b%MOD)) % MOD;
	return (ret+MOD)%MOD;
}
 
int add(int a, int b){
	int ret = (1ll * (a%MOD) + (b%MOD)) % MOD;
	return (ret+MOD)%MOD;
}
 
int pow_exp(int n, int p){
	if(!p) return 1;
	if(p&1) return mul(n, pow_exp(n, p-1));
	int tmp = pow_exp(n, p/2);
	return mul(tmp, tmp);
}

int inv(int x){
	return pow_exp(x, MOD-2);
}
 
const int  MAX = 2e5 + 10;
const int NMAX = 2e5 + 10;
const int MMAX = 2e5 + 10;
const int LOG_MAX = ceil(log2(double(NMAX)));
const int BLOCK = ceil(sqrt(double(NMAX)));

int fac[NMAX], ifac[NMAX];

void init(){
	
	fac[0] = 1;
	for(int i = 1; i < NMAX; i++){
		fac[i] = mul(fac[i-1], i);
	}
	
	ifac[NMAX-1] = inv(fac[NMAX-1]);
	for(int i = NMAX-2; i >= 0; i--){
		ifac[i] = mul(ifac[i+1], i+1);
	}
}

int choose(int n, int c){
	assert(n >= c);
	return mul(fac[n], mul(ifac[c], ifac[n-c]));
}

typedef long double dt;


struct segTree{
	
	vi tree;
	
	segTree(int n){
		tree = vi(n*4);
	}
	 
	void update(int nd, int l, int r, int p, int v){
		
		if(p < l || r < p){
			return;
		}
		
		if(l==r){
			tree[nd] = max(tree[nd], v);
			return;
		}
		
		int mid = (l+r)/2;
		update(nd*2, l, mid, p, v);
		update(nd*2+1, mid+1, r, p, v);
		
		tree[nd] = max(tree[nd*2], tree[nd*2+1]);
	}
	
	int query(int nd, int l, int r, int q_l, int q_r){
		
		if(r < q_l || q_r < l){
			return 0;
		}
		
		if(q_l <= l && r <= q_r){
			return tree[nd];
		}
		
		int mid = (l+r)/2;
		int st_path = query(nd*2, l, mid, q_l, q_r);
		int nd_path = query(nd*2+1, mid+1, r, q_l, q_r);
		
		return max(st_path, nd_path);
	}
};


int32_t main(){
    
    fastIO;
    
#ifdef LOCAL
    freopenI;
    freopenO;
#endif

    // freopen("name.in", "r", stdin);
    
	// init();
	
	int n, k;
	cin >> n >> k;
	
	vpii vec(n);
	for(auto &[x, y] : vec) cin >> x >> y;
	
	auto check = [&](dt m){
		
		vi tmp(n);
		for(int i = 0; i < n; i++){
			auto [x, y] = vec[i];
			dt v = dt(y)-m*dt(x);
			tmp[i] = int(v * dt(1e9));
		}
		
		vi ttmp = tmp;
		sort(ttmp.begin(), ttmp.end());
		ttmp.erase(unique(ttmp.begin(), ttmp.end()), ttmp.end());
		
		for(auto &e : tmp){
			e = lower_bound(ttmp.begin(), ttmp.end(), e) - ttmp.begin();
		}
		
		int sz = ttmp.size();
		
		segTree st(sz);
		for(int i = 0; i < n; i++){
			int v = tmp[i];
			int cans = st.query(1, 0, sz-1, 0, v) + 1;
			st.update(1, 0, sz-1, v, cans);
		}
		
		return st.query(1, 0, sz-1, 0, sz-1);
	};
	
	dt l = -1e10;
	dt r = 1e10;
	
	dt ans = -1;
	
	int cnt = 200;
	while(cnt--){
		
		dt mid = l + (r-l)/2.0;
		
		if(check(mid) >= k){
			ans = mid;
			l = mid;
		}
		else{
			r = mid;
		}
	}
	
	cout << fixed << setprecision(10) << ans << endl;
}

详细

Test #1:

score: 100
Accepted
time: 1ms
memory: 3808kb

input:

4 3
1 2
2 4
3 3
4 1

output:

-0.9999999997

result:

ok found '-1.0000000', expected '-1.0000000', error '0.0000000'

Test #2:

score: 0
Accepted
time: 1ms
memory: 3856kb

input:

2 2
1 1
5 3

output:

0.5000000002

result:

ok found '0.5000000', expected '0.5000000', error '0.0000000'

Test #3:

score: -100
Wrong Answer
time: 1ms
memory: 3744kb

input:

2 2
222640995 547139825
489207317 725361095

output:

10000000000.0000000000

result:

wrong answer 1st numbers differ - expected: '0.6685813', found: '10000000000.0000000', error = '9999999999.3314190'