Codeforces 1097 Alex and a TV Show

传送门
除了操作 \(3\) 都可以 \(bitset\)
现在要维护
\[C_i=\sum_{gcd(j,k)=i}A_jB_k\]
类比 \(FWT\)，只要求出 \(A'_i=\sum_{i|d}A_d\)
就可以直接按位相乘了
求答案就是莫比乌斯反演，\(A_i=\sum_{i|d}\mu(\frac{d}{i})A'_i\)
把每个数字的 \(\mu\) 的 \(bitset\) 预处理出来，乘法就是 \(and\)
最后用 \(count\) 统计答案

# include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn(7005);

int mu[maxn], n, q, pr[maxn], tot;
bitset <maxn> bc[100005], bcm[7005], bcv[7005], ispr;

int main() {
    int i, j, op, l, r, v;
    mu[1] = 1;
    for (i = 2; i <= 7000; ++i) {
        if (!ispr[i]) pr[++tot] = i, mu[i] = -1;
        for (j = 1; j <= tot && pr[j] * i <= 7000; ++j) {
            ispr[pr[j] * i] = 1;
            if (i % pr[j]) mu[i * pr[j]] = -mu[i];
            else {
                mu[i * pr[j]] = 0;
                break;
            }
        }
    }
    for (i = 1; i <= 7000; ++i)
        for (j = i; j <= 7000; j += i) {
            bcv[j].set(i);
            if (mu[j / i]) bcm[i].set(j);
        }
    scanf("%d%d", &n, &q);
    for (i = 1; i <= q; ++i) {
        scanf("%d%d%d", &op, &l, &r);
        if (op == 1) bc[l] = bcv[r];
        else if (op == 2) scanf("%d", &v), bc[l] = bc[r] ^ bc[v];
        else if (op == 3) scanf("%d", &v), bc[l] = bc[r] & bc[v];
        else putchar(((bc[l] & bcm[r]).count() & 1) + '0');
    }
    return 0;
}