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2021杭电多校2 K I love max and multiply

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题目描述

https://acm.hdu.edu.cn/showproblem.php?pid=6971

简要题意：给定两个长度为 $n$ 的序列 $A_i$ 和 $B_i$，求另一个长度为 $n$ 的序列 $C_i$，满足 $C_k=\max\lbrace A_iB_j\rbrace$，其中 $i~and~j\ge k$

$n< 2^{18}$

Solution

比较直接的想法是求 $D_k=\max_{i~and~j=k}A_iB_j$，然后倒着做一遍 $\max$

但是这个东西并不好求，所以我们还是回归到原问题 $C_k=\max_{i~and~j\ge k}A_iB_j$

我们考虑求 $E_k=\max_{k\in i~and~j}A_iB_j$，最后倒着做一遍 $\max$ 即可

这个东西显然可以直接上高维前缀和

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#include <iostream>
#define maxn 400010
#define ll long long
#define INF 1000000001
#define inf 1000000000000000001
using namespace std;

const int p = 998244353;

int n, A[maxn], B[maxn], a[maxn], b[maxn];

void work() {
    cin >> n;
    for (int i = 0; i < n; ++i) cin >> a[i], A[i] = a[i];
    for (int i = 0; i < n; ++i) cin >> b[i], B[i] = b[i];
    int m = 1;
    while (m < n) m <<= 1;
    for (int i = n; i < m; ++i) A[i] = B[i] = -INF, a[i] = b[i] = INF;
    n = m;
    for (int o = 0; (1 << o) < n; ++o)
        for (int S = n - 1; ~S; --S)
            if (!(S >> o & 1)) {
                A[S] = max(A[S], A[S | 1 << o]);
                B[S] = max(B[S], B[S | 1 << o]);
                a[S] = min(a[S], a[S | 1 << o]);
                b[S] = min(b[S], b[S | 1 << o]);
            }
    ll ans = -inf, sum = 0; 
    for (int i = n - 1; ~i; --i) {
        if (A[i] != -INF && B[i] != -INF) ans = max(ans, 1ll * A[i] * B[i]);
        if (A[i] != -INF && b[i] != INF) ans = max(ans, 1ll * A[i] * b[i]);
        if (a[i] != INF && B[i] != -INF) ans = max(ans, 1ll * a[i] * B[i]);
        if (a[i] != INF && b[i] != INF) ans = max(ans, 1ll * a[i] * b[i]);
        if (ans != -inf) sum = (sum + ans) % p; 
    } cout << (sum + p) % p << "\n"; 
}

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

    int T; cin >> T;
    while (T--) work(); 
    return 0;
}