34 lines
855 B
C++
34 lines
855 B
C++
// 102
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// https://www.luogu.com.cn/article/42ly2wrn
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#include <bits/stdc++.h>
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#define int long long
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using namespace std;
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int n, l;
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int sum[100005];
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int q[100005];
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int t, h;
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double calc(int x, int y)
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{
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return (0.0 + sum[y] - sum[x]) / (y - x);
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}
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signed main()
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{
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scanf("%lld%lld", &n, &l);
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for (int i = 1, x; i <= n; ++i)
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scanf("%lld", &x), sum[i] = sum[i - 1] + x;
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double ans = 0;
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for (int i = l; i <= n; ++i) // lp
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{
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while (t - h >= 2 && calc(i - l, q[t - 1]) < calc(i - l, q[t - 2]))
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t--;
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q[t++] = i - l;
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while (t - h >= 2 && calc(i, q[h]) < calc(i, q[h + 1]))
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h++;
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ans = max(ans, calc(i, q[h]));
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}
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printf("%lld\n", (long long)floor(1000 * ans));
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return 0;
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}
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/*
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连续子序列平均值就转化为S-x平面上的斜率：avg(x,y)=(S(y)-s(x-1))/(y-x+1)
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*/ |