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;
}