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 
    
     using namespace std;typedef long long ll;const int maxn(7005);int mu[maxn], n, q, pr[maxn], tot;bitset 
     
       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;}