luogu1419 寻找段落 （二分，单调队列）

单调队列存坐标

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#define R(a,b,c) for(register int  a = (b); a <= (c); ++ a)

#define nR(a,b,c) for(register int  a = (b); a >= (c); -- a)

#define Max(a,b) ((a) > (b) ? (a) : (b))

#define Min(a,b) ((a) < (b) ? (a) : (b))

#define Fill(a,b) memset(a, b, sizeof(a))

#define Abs(a) ((a) < 0 ? -(a) : (a))

#define Swap(a,b) a^=b^=a^=b

#define ll long long

#define ON_DEBUG

#ifdef ON_DEBUG

#define D_e_Line printf("\n\n----------\n\n")

#define D_e(x)  cout << #x << " = " << x << endl

#define Pause() system("pause")

#define FileOpen() freopen("in.txt","r",stdin);

#else

#define D_e_Line ;

#define D_e(x)  ;

#define Pause() ;

#define FileOpen() ;

#endif

struct ios{

    template<typename ATP>ios& operator >> (ATP &x){

        x = 0; int f = 1; char c;

        for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-')  f = -1;

        while(c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();

        x*= f;

        return *this;

    }

}io;

using namespace std;

const int N = 100007;

int n, lenS, lenT;

int a[N];

double b[N], sum[N];

inline bool Check(double mid){

    R(i,1,n){

    	sum[i] = sum[i - 1] + (double)a[i] - mid;

    }

    int *q = new int[n + 3];

    int h = 1, t = 0;

    R(i,lenS,n){

        while(h <= t && sum[i - lenS] < sum[q[t]]) --t;

        q[++t] = i - lenS;

        if(h <= t && q[h] < i - lenT) ++h;

        if(h <= t && sum[i] - sum[q[h]] >= 0){

        	delete []q;

        	return true;

        }

    }

    delete []q;

    return false;

}

int main(){

	io >> n >> lenS >> lenT;

	R(i,1,n){

		io >> a[i];

	}

	double l = -10000, r = 10000;

	while(r - l > 1e-6){

		double mid = (l + r) / 2.0;

		if(Check(mid))

			l = mid;

		else

			r = mid;

	}

	printf("%.3lf", l);

	return 0;

}

luogu1419 寻找段落 （二分，单调队列）的相关教程结束。