// Problem: B. Composite Coloring
// Contest: Codeforces - Codeforces Round #630 (Div. 2)
// URL: https://codeforces.com/problemset/problem/1332/B
// Memory Limit: 512 MB
// Time Limit: 2000 ms
// 2022-03-02 15:11:35
//
// Powered by CP Editor (https://cpeditor.org)

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

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

const ll mod = 1e9 + 7;

inline ll qmi (ll a, ll b) {
ll ans = 1;
while (b) {
if (b & 1) ans = ans * a%mod;
a = a * a %mod;
b >>= 1;
}
return ans;
}
inline int read () {
int x = 0, f = 0;
char ch = getchar();
while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
return f?-x:x;
}
template<typename T> void print(T x) {
if (x < 0) putchar('-'), x = -x;
if (x >= 10) print(x/10);
putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
return (a + b) %mod;
}
inline ll inv (ll a) {
return qmi(a, mod - 2);
}
const int N = 1e5 + 10;
int n;
int a[N];
int ans[N];
vector<int> primes;
int vis[N];
int gcd (int a, int b) {
return b?gcd(b, a %b) : a;
}
bool check (int tar) {
if (tar < 2) return 0;
for (int i =2; i <= tar / i; i ++)
if (tar % i == 0) return 0;
return 1;
}
void solve() {

cin >> n;

for (int i = 1; i <= n ;i ++)
cin >> a[i];
memset(vis, 0, sizeof vis);
int cnt = 0;
for (int i = 1; i <= n ;i ++) {
for (int j = 0; j < primes.size(); j ++)
if (a[i] % primes[j] == 0) {
if (vis[primes[j]])
ans[i] = vis[primes[j]];
else
ans[i] = vis[primes[j]] = ++cnt;
break;
}

}
cout << cnt << endl;
for (int i = 1; i <= n; i ++) cout << ans[i] << " ";
puts("");
}
int main () {

for (int i = 2; i <= 31; i ++)
if (check(i)) primes.pb(i);
int t;
t =1;
cin >> t;
while (t --) solve();
return 0;
}


通过素数分解对数进行分类 sqrt(1000)刚好只涉及11个素数. 然后实现细节由于需要连续,所以我们需要cnt变量把涉及到的素数从小到大标记