P1419 尋找段落

題目連接

思路：

首先二分答案，即：二分最大平均值。

咱們將a所有減去mid，問題轉化爲判斷是否存在一個長度在s~t範圍內的區間它的和爲正，若是有說明還有更大的平均值。

用前綴和和單調隊列維護。

而後用單調隊列求出sum[i]-min(sum[i-t]~sum[i-s]),而後判斷是否大於0便可。

Code

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 1e5 + 5;

int n;
int s,t;
int a[MAXN];
double sum[MAXN];

inline int read() {
    int k = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
        if (ch == '-')
            f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
        k = (k << 1) + (k << 3) + (ch & 15);
        ch = getchar();
    }
    if (f == -1) k = ~k + 1;
    return k;
}    

int head = 1,tail = 0,q[MAXN];

bool check(double mid){
    memset(q,0,sizeof(q));
    memset(sum,0,sizeof(sum));
    
    for(int i = 1 ; i <= n ; i++)   sum[i] = sum[i - 1] + double(a[i]) - mid;
    
    head = 1,tail = 0;
    
    for(int i = s ; i <= n ; i++){
        while(head <= tail && sum[q[tail]] > sum[i - s]) tail--;
        q[++tail] = i - s;
        while(head <= tail && q[head] < i - t) head++;
        if(head <= tail && sum[i] - sum[q[head]] >= 0) return true;
    }
    
    return false;
}

int main(){
    n = read(),s = read(),t = read();

    for (int i = 1 ; i <= n ; i++) a[i] = read();

    double l = -10000.0,r = 10000.0;

    while(r - l > 1e-4){
        double mid = (l + r) / 2.0;
        if (check(mid) == true) l = mid;
        else r = mid;
    }

    printf("%.3lf",l);
    return 0;
}