hdu-6579 Operation

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题目链接

Operation

Problem Description

There is an integer sequence a of length n and there are two kinds of operations:

• 0 l r: select some numbers from al...ar so that their xor sum is maximum, and print the maximum value.

• 1 x: append x to the end of the sequence and let n=n+1.

Input

There are multiple test cases. The first line of input contains an integer T(T≤10), indicating the number of test cases.
For each test case:
The first line contains two integers n,m\((1≤n≤5×10^5,1≤m≤5×10^5)\), the number of integers initially in the sequence and the number of operations.
The second line contains n integers \(a1,a2,...,an(0≤ai<2^{30})\), denoting the initial sequence.
Each of the next m lines contains one of the operations given above.
It‘s guaranteed that \(∑n≤10^6,∑m≤10^6,0≤x<2^{30}\).
And operations will be encrypted. You need to decode the operations as follows, where lastans denotes the answer to the last type 0 operation and is initially zero:
For every type 0 operation, let l=(l xor lastans)mod n + 1, r=(r xor lastans)mod n + 1, and then swap(l, r) if l>r.
For every type 1 operation, let x=x xor lastans.

Output

For each type 0 operation, please output the maximum xor sum in a single line.

Sample Input

1
3 3
0 1 2
0 1 1
1 3
0 3 4

Sample Output

1
3

题意

给一个长度为n的数组m个操作

• 0 x y 查询区间\([x,y]\)取任意个数能异或出的最大值
• 1 x 向数组尾部添加一个数x

强制在线

题解

朴素的线性基只能查询1-n能异或出的最大值，这题我们可以保存\([1,n]\)每个前缀线性基的状态,查询x,y时只需要查询第y个前缀的线性基就行
但是前缀里会有1-x的线性基影响结果，我们可以在插入线性基时做处理，如果在第pos位上已经有数，且这个数的插入时间比我当前数的插入时间早，那么就把当前要插入的数与该数交换，当前插入时间也交换，直至当前数无法插入或变为0
这样可以让前缀线性基里的数都是越新的，查询的时候判断线性基上数的插入时间是否大于等于x，如果大于x就可以使用这个数。这样处理的正确性是因为线性基插入不受顺序影响，同一组数以不同顺序插入，最后得到的线性基都是等价的

代码

#include <bits/stdc++.h>

const int mx = 1e6+5;
typedef long long ll;

int sum[mx][32];
int pos[mx][32];
int tot;

void add(int num) {
    ++tot;
    for (int i = 0; i < 32; i++) {
        sum[tot][i] = sum[tot-1][i];
        pos[tot][i] = pos[tot-1][i];
    }

    int now = tot;
    for (int i = 30; i >= 0; i--) {
        if (num & (1<<i)) {
            if (sum[tot][i] == 0) {
                sum[tot][i] = num;
                pos[tot][i] = now;
                break;
            }

            if (now > pos[tot][i]) {
                std::swap(now, pos[tot][i]);
                std::swap(num, sum[tot][i]);  
            }
            num ^= sum[tot][i];
        }
    }
}

int query(int l, int r) {
    int ans = 0;
    for (int i = 30; i >= 0; i--) {
        if (sum[r][i] && pos[r][i] >= l) {
            ans = std::max(ans, ans ^ sum[r][i]);
        }
    }
    return ans;
}

int main() {
    int T;
    scanf("%d", &T);

    while (T--) {
        int lastans = 0; tot = 0;
        int n, m, num;
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= n; i++) {
            scanf("%d", &num);
            add(num);
        }

        while (m--) {
            int op, l, r;
            scanf("%d", &op);
            if (op == 0) {
                scanf("%d%d", &l, &r);
                l = (l ^ lastans) % n + 1;
                r = (r ^ lastans) % n + 1;
                if (l > r) std::swap(l, r);
                lastans = query(l, r);
                printf("%d\n", lastans);
            } else {
                scanf("%d", &r);
                add(r ^ lastans);
                n++;
            }
        }
    }
    return 0;
}

hdu-6579 Operation

标签：answer   nta   ica   题意   initial   不同   eve   最大   out   

原文地址：https://www.cnblogs.com/bpdwn-cnblogs/p/11290776.html