Cod sursa(job #1301880)

Utilizator mirceadinoMircea Popoveniuc mirceadino Data 26 decembrie 2014 14:20:01
Problema Secventa 3 Scor 100
Compilator cpp Status done
Runda Arhiva de probleme Marime 2.17 kb
#include<algorithm>
#include<bitset>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<deque>
#include<fstream>
#include<iomanip>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<unordered_map>
#include<unordered_set>
#include<utility>
#include<vector>

using namespace std;

#define dbg(x) (cout<<#x<<" = "<<(x)<<'\n')
#ifdef HOME
const string inputFile = "input.txt";
const string outputFile = "output.txt";
#else
const string problemName = "secv3";
const string inputFile = problemName + ".in";
const string outputFile = problemName + ".out";
#endif

typedef long long int lld;
typedef pair<int, int> PII;
typedef pair<int, lld> PIL;
typedef pair<lld, int> PLI;
typedef pair<lld, lld> PLL;

const int INF = (1LL << 31) - 1;
const lld LINF = (1LL << 62) - 1;
const int dx[] = {1, 0, -1, 0, 1, -1, 1, -1};
const int dy[] = {0, 1, 0, -1, 1, -1, -1, 1};
const double EPS = (1e-3);
const int NMAX = 100000 + 5;

int N, L, U;
int C[NMAX];
int T[NMAX];
double A[NMAX];
double S[NMAX];
double sol;
deque<int> DQ;

bool check(double x) {
    int i;

    for(i = 1; i <= N; i++) {
        A[i] = C[i] - x * T[i];
        S[i] = S[i - 1] + A[i];
    }

    DQ.resize(0);

    for(i = L; i <= N; i++) {
        while(!DQ.empty() && S[i - L] <= S[DQ.back()])
            DQ.pop_back();

        while(!DQ.empty() && i - DQ.front() + 1 > U)
            DQ.pop_front();

        DQ.push_back(i - L);

        if(S[i] - S[DQ.front()] >= 0)
            return 1;
    }

    return 0;
}

int main() {
    int i;
    double lo, mi, hi;

#ifndef ONLINE_JUDGE
    freopen(inputFile.c_str(), "r", stdin);
    freopen(outputFile.c_str(), "w", stdout);
#endif

    scanf("%d%d%d", &N, &L, &U);

    for(i = 1; i <= N; i++)
        scanf("%d", &C[i]);

    for(i = 1; i <= N; i++)
        scanf("%d", &T[i]);

    for(lo = 0.0, hi = 10000.0; fabs(lo - hi) > EPS; ) {
        mi = (lo + hi) / 2.0;
        if(check(mi)) {
            lo = mi;
            sol = mi;
        } else
            hi = mi;
    }

    printf("%.2f\n", sol);

    return 0;
}