I love max and multiply

题目链接

题意:给定两数组$A$,$B$,定义$C$数组为:$C_k=max(A_i\ast B_j)$,$i$&$j$ $\ge k$,求${\sum }_{i=1}^n{C}_i$$modp$

解析:我们定义$D$数组为:$D_k=max(A_i\ast B_j,D_{k+1})$,$i$&$j=k$,则$C_i=D_i$,我们从$k=n$开始求$D_k$,每次用四个数组$maxa,mina,maxb,minb$记录和$i$进行与操作后可以大于等于$i$的所有下标的最大$A$,最小$A$,最大$B$,最小$B$。
接下来感性理解一下:以$maxa$进行举例: $max{a}_i=max(max{a}_{i|(1<<j)},A_i),i|(1<<j)\le n-1$,$i|(1<<j)$可以得到至多将1个0变为1的的最大A,而$max{a}_{i|1<<j}$得到的是$i|(1<<j)$至多将1个0变为1的的最大A,以此类推,$max{a}_i$可以得到所有$i$&$j$$\ge i$的A。其他三数组同理。
因为$|A|,|B|\le 1e9$,为了防止负数,所以需要$mina,minb$来处理负数情况,$C_i=max(C_{i+1},maxa\ast maxb,mina\ast minb,maxa\ast minb,maxb\ast mina)$

AcCode:

#include <iostream>
#include <algorithm>
#include <vector>
#include <map>

#define inf 2e18
#define int long long

const int N = 1e6 + 100;
const int mod = 998244353;

int a[N], b[N], maxa[N], maxb[N], mina[N], minb[N];

inline int max(int a, int b) { return (a > b) ? a : b; }
inline int min(int a, int b) { return (a < b) ? a : b; }
inline int max(int a, int b, int c, int d, int e) {
    int res = a;
    if (b > res) res = b;
    if (c > res) res = c;
    if (d > res) res = d;
    if (e > res) res = e;
    return res;
}

signed main() {
    int t; scanf("%lld", &t);
    const int sign = 1;
    while (t--) {
        int n; scanf("%lld", &n);
        for (int i = 0; i < n; i++) scanf("%lld", &a[i]), maxa[i] = mina[i] = a[i];
        for (int i = 0; i < n; i++) scanf("%lld", &b[i]), maxb[i] = minb[i] = b[i];
        for (int i = n - 1; i >= 0; i--) {
            for (int j = 0; (sign << j) < n; j++) {
                int temp = (sign << j);
                temp = temp | i;
                if (temp >= n) continue;
                maxa[i] = max(maxa[i], maxa[temp]);
                mina[i] = min(mina[i], mina[temp]);
                maxb[i] = max(maxb[i], maxb[temp]);
                minb[i] = min(minb[i], minb[temp]);
            }
        }
        int ans = 0, rem = -inf;
        for (int i = n - 1; i >= 0; i--) {
            rem = max(rem, maxa[i] * maxb[i], mina[i] * minb[i], maxa[i] * minb[i], mina[i] * maxb[i]);
            ans = (ans + rem) % mod;
        }
        printf("%lld\n", (ans % mod + mod) % mod);
    }
}