CF212D 【Cutting a Fence】

AC传送门！

由于在洛谷发布时，用了大量的LATEX,所以不想在博客园重新打一遍了。

洛谷博客

#include <bits/stdc++.h>
using namespace std;
#define N 1000010
#define ll long long

inline ll read(){
    ll x = 0, s = 1;
    char c = getchar();
    while(!isdigit(c)){
        if(c == '-')s = -1;
        c = getchar();
    }
    while(isdigit(c)){
        x = x * 10 + (c ^ '0');
        c = getchar();
    }
    return x * s;
}

ll n, m, r[N], l[N], a[N];
ll s1[N], s2[N];
double ans[N];

void search(){
    for(int i = n; i; i--){  /*向右找到第一个小于等于这个数的位置*/
        if(a[i + 1] <= a[i]) r[i] = i + 1;
        else 
            for(int j = r[i + 1]; j <= n + 1; j = r[j])  /*一直向右跑*/
                if(a[j] <= a[i]){
                    r[i] = j;
                    break;
                }
    }
    for(int i = 1;i <= n; i++){  /*同上，向左找*/
        if(a[i - 1] < a[i]) l[i] = i - 1;
        else 
            for(int j = l[i - 1]; j >= 0; j = l[j]) 
                if(a[j] < a[i]) {
                    l[i] = j;
                    break;
                }

    }
    for(int i = 1;i <= n; i++){  /*换算成距离 L 和 R*/
        l[i] = i - l[i];   
        r[i] = r[i] - i;
    }
    return ;
}

void prepare(){  /*进行差分以及预处理答案*/
        for(int i = 1;i <= n; i++){
        int x = l[i], y = r[i];
        if(x > y) swap(x, y);  /*保持 L < S*/
        s1[1] += a[i];
        s1[x + 1] -= a[i];
        s2[x + 1] += a[i] * x;
        s1[y + 1] -= a[i];
        s2[y + 1] += a[i] * y;
        s1[x + y] += a[i]; 
        s2[x + y] -= a[i] * (x + y);
    }
    ll num1 = 0, num2 = 0;
    for(int i = 1;i <= n; i++){
        num1 += s1[i], num2 += s2[i];
        ans[i] = ((num1 * i + num2) * 1.00) / ((n - i + 1) * 1.00); /*预处理出所有答案*/
    }
    return ;
}

int main(){
    n = read();
    for(int i = 1;i <= n; i++)
        a[i] = read();
    a[n + 1] = (ll)(-(1 << 50));/*两端插入极小值，避免下一步越界或者出现错误*/
    a[0] = (ll)(-(1 << 50));
    search(); /*先搜索距离*/
    prepare();  /*然后处理答案*/
    m = read();
    while(m--){
        ll k = read();
        printf("%.9lf
", ans[k]);  
    }
    return 0;
}