• HDU6579 Operation(线性基)


    题意:

    有两种操作:

    • 0 l r 在区间l,r里面选择一些数,使得他们的异或和最大,并输出这个最大值。
    • 1 x 将x推进序列,序列长度加1。

    题解:

    看了网上的线性基才会做,知识盲区...

    #include<cstdio>
    #include<algorithm>
    #include<cstring>
    using namespace std;
    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]) {
                    swap(now, pos[tot][i]);
                    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) swap(l, r);
                    lastans = query(l, r);
                    printf("%d
    ", lastans);
                } else {
                    scanf("%d", &r);
                    add(r ^ lastans);
                    n++;
                }
            }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zhanglichen/p/12493718.html
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