[HDU 3949] XOR

Posted 2021-01-22 fly-in-milkyway

Description

求 (n) 个数的第 (k) 小的异或和。

Solution

把 (k) 二进制拆分，若第 (i) 位上是 (1)，就把线性基中第 (i) 个元素异或起来。

注意要维护非重叠的右上角矩阵。

Code

#include <cstdio>
#include <cstring>

typedef long long LL;
int T, n, m, cnt; LL a[10005], p[65], q[65], k, x, ans;

void insert(LL x) {
    for (int i = 62; ~i; --i) {
        if (!(x >> i)) continue;
        if (!p[i]) { p[i] = x; break; }
        x ^= p[i];
    }
}
void rebuild() {
    for (int i = 62; ~i; --i)
        for (int j = i - 1; ~j; --j)
            if (p[i] & (1LL << j)) p[i] ^= p[j];
    for (int i = 0; i <= 62; ++i)
        if (p[i]) q[++cnt] = p[i];
}

int main() {
    scanf("%d", &T);
    for (int cas = 1; cas <= T; ++cas) {
        printf("Case #%d:
", cas);
        memset(p, 0, sizeof p);
        scanf("%d", &n), cnt = 0;
        for (int i = 1; i <= n; ++i) scanf("%lld", &x), insert(x);
        rebuild();
        scanf("%d", &m);
        while (m--) {
            scanf("%lld", &k), ans = 0;
            if (cnt == n) ++k;
            if (k > (1LL << cnt)) { puts("-1"); continue; }
            if (--k == 0) { puts("0"); continue; }
            int j = 1;
            while (k) {
                if (k & 1) ans ^= q[j];
                k >>= 1, ++j;
            }
            printf("%lld
", ans);
        }
    }
    return 0;
}