14318: 【原4318】最大平均子区间

题目

题目描述

author: smallling 原OJ链接：https://acm.sjtu.edu.cn/OnlineJudge-old/problem/4318

Description

给定一个长度为$n$的数组$a$，求$\max\limits_{1\leq l \leq r \leq n}\left\lfloor\frac{\sum_{i=l}^{r}a_i}{r-l+1}\right\rfloor$

Input Format

第一行一个数字$n$表示数组长度。 接下来一行$n$个数字表示数组$a$。 $1 \leq n \leq 10^6, 1 \leq a_{i} \leq 10^{9}$

Output Format

一行一个数字表示答案

Sample Input

9
3 5 3 7 7 6 6 5 6

Sample Output

7

ligongzzz's solution

#include <iostream>
#include <algorithm>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    int n;
    cin >> n;

    int max_n = 0;
    for (int i = 0; i < n; ++i) {
        int temp;
        cin >> temp;
        max_n = max(temp, max_n);
    }
    cout << max_n;
    return 0;
}

zqy2018's solution

#include <bits/stdc++.h>
#define REP(temp, init_val, end_val) for (int temp = init_val; temp <= end_val; ++temp)
using namespace std;
typedef long long ll;
int read(){
    int f = 1, x = 0;
    char c = getchar();
    while (c < '0' || c > '9'){if(c == '-') f = -f; c = getchar();}
    while (c >= '0' && c <= '9')x = x * 10 + c - '0', c = getchar();
    return f * x; 
}
int n, a[1000005], maxi;
void init(){
    n = read();
    maxi = 0;
    REP(i, 1, n)
        a[i] = read(), maxi = max(maxi, a[i]);
}
ll get(int x){
    ll mini = 0;
    ll ans = -10000000000000000ll, sm = 0;
    REP(i, 1, n){
        sm += a[i] - x;
        ans = max(ans, sm - mini);
        mini = min(mini, sm);
    }
    return ans;
}
void solve(){
    int l = 0, r = maxi;
    while (r > l){
        int mid = (r + l + 1) >> 1;
        if (get(mid) >= 0ll)
            l = mid;
        else r = mid - 1;
    }
    printf("%d\n", l);
}
int main(){
    init();
    solve();
    return 0;
}