bzoj 5090 组题

传送门

两种方法做这道题。

1、二分答案。

2、因为所求的表达式是一个斜率的形式，所以维护一个凸包。然后对于每一个右端点在凸包上找一个使其最大的值就可以了。

#include<cstdio>
#include<iostream>
using namespace std;
inline char nc() {
    static char b[100000],*s=b,*t=b;
    return s==t&&(t=(s=b)+fread(b,1,100000,stdin),s==t)?-1:*s++;
}
template < class T >
inline void read(T &x) {
    char b = nc(); x = 0; bool f = 0;
    for (; !isdigit(b); b = nc()) if (b == '-') f = 1;
    for (; isdigit(b); b = nc()) x = x * 10 + b - '0'; if (f) x = -x;
}
typedef long long ll;
typedef double db;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
const int N = 1000010;
int n, k, z[N], top, x, y;
ll s[N], P, Q, G;
db tans;
bool judge(int a, int b, int c) {
    return (s[b] - s[a]) * (c - b) > (s[c] - s[b]) * (b - a);
}
db calc(int a, int b) {
    return db(s[b] - s[a]) / (b - a);
}
void upd(int p) {
    int l = 1, r = top, ll, rr;
    while (r - l > 3) {
        ll = (r - l + 1) / 3 + l;
        rr = (r - l + 1) / 3 * 2 + l;
        calc(z[ll], p) < calc(z[rr], p) ? l = ll : r = rr;
    }
    for (ll = l++; l <= r; ++l)
        if(calc(z[l], p) > calc(z[ll], p)) ll = l;
    db res = db(s[p] - s[z[ll]]) / (p - z[ll]);
    if (res > tans) tans = res, x = z[ll], y = p;
}
int main() {
    read(n); read(k); tans = -1e9;
    for (int i = 1; i <= n; ++i) read(s[i]), s[i] += s[i-1];
    for (int i = 0; i <= n - k; ++i) {
        while (top > 1 && judge(z[top-1], z[top], i)) --top;
        z[++top] = i; upd(i + k);
    } P = s[y] - s[x]; Q = y - x; G = gcd(P, Q);
    if (G < 0) G = -G; P /= G; Q /= G;
    printf("%lld/%lld\n", P, Q);
    return 0;
}
/*
int n, k, pb[N];
ll a[N], sa[N], P, Q, G;
db b[N];
bool check(db m) {
    for (int i = 1; i <= n; ++i) {
        b[i] = b[i-1] + a[i] - m;
        pb[i] = (b[i] < b[pb[i-1]]) ? i : pb[i-1];
    }
    for (int i = k; i <= n; ++i) {
        if (b[i] - b[pb[i-k]] > 0)
            return P = sa[i] - sa[pb[i-k]], Q = i - pb[i-k], true;
    } return false;
}
int main() {
    read(n); read(k);
    for (int i = 1; i <= n; ++i) read(a[i]);
    for (int i = 1; i <= n; ++i) sa[i] = sa[i-1] + a[i];
    db l = -1e8, r = 1e8;
    for (int i = 0; i < 100; ++i) {
        db m = (l + r) / 2;
        check(m) ? l = m : r = m;
    } G = gcd(P, Q); if (G < 0) G = -G;
    if (G) P /= G, Q /= G;
    printf("%lld/%lld\n", P, Q);
    return 0;
}
*/