题目传送门

一、两层暴力循环法

#include <bits/stdc++.h>

using namespace std;
const int N = 3010;
const int INF = 0x3f3f3f3f;
typedef long long LL;
int a[N];
LL MIN = INF;

int main() {
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> a[i];

    for (int i = 1; i <= n - m + 1; i++) {
        LL sum = 0;
        for (int j = 0; j < m; j++) sum += a[i + j];
        MIN = min(MIN, sum);
    }
    cout << MIN << endl;
    return 0;
}

二、滑动窗口法

#include <bits/stdc++.h>

using namespace std;
const int N = 3010;
const int INF = 0x3f3f3f3f;
int a[N];
int MIN = INF;

int main() {
    //滑动窗口思想
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> a[i];

    //第一组m个数
    int sum = 0;
    for (int i = 1; i <= m; i++) sum += a[i];
    MIN = sum;//猴子选大王

    //滑啊滑啊滑
    for (int i = m + 1; i <= n; i++) {
        sum += a[i] - a[i - m];
        MIN = min(MIN, sum);
    }
    //输出大吉！
    cout << MIN << endl;
    return 0;
}