A positive integer is called composite if it can be represented as a product of two positive integers, both greater than 1. For example, the following numbers are composite: 6, 4, 120, 27. The following numbers aren’t: 1, 2, 3, 17, 97.
Alice is given a sequence of n composite numbers a1,a2,…,an.
She wants to choose an integer m≤11 and color each element one of m colors from 1 to m so that:
for each color from 1 to m there is at least one element of this color;
each element is colored and colored exactly one color;
the greatest common divisor of any two elements that are colored the same color is greater than 1, i.e. gcd(ai,aj)>1 for each pair i,j if these elements are colored the same color.
Note that equal elements can be colored different colors — you just have to choose one of m colors for each of the indices from 1 to n.
Alice showed already that if all ai≤1000 then she can always solve the task by choosing some m≤11.
Help Alice to find the required coloring. Note that you don’t have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Input
The first line contains a single integer t (1≤t≤1000) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1≤n≤1000) — the amount of numbers in a sequence a.
The second line of the test case contains n composite integers a1,a2,…,an (4≤ai≤1000).
It is guaranteed that the sum of n over all test cases doesn’t exceed 104.
Output
For each test case print 2 lines. The first line should contain a single integer m (1≤m≤11) — the number of used colors. Consider colors to be numbered from 1 to m. The second line should contain any coloring that satisfies the above conditions. Print n integers c1,c2,…,cn (1≤ci≤m), where ci is the color of the i-th element. If there are multiple solutions then you can print any of them. Note that you don’t have to minimize or maximize the number of colors, you just have to find the solution with some m from 1 to 11.
Remember that each color from 1 to m should be used at least once. Any two elements of the same color should not be coprime (i.e. their GCD should be greater than 1).
Example
input
3
3
6 10 15
2
4 9
23
437 519 865 808 909 391 194 291 237 395 323 365 511 497 781 737 871 559 731 697 779 841 961
output
1
1 1 1
2
2 1
11
4 7 8 10 7 3 10 7 7 8 3 1 1 5 5 9 2 2 3 3 4 11 6
Note
In the first test case, gcd(6,10)=2, gcd(6,15)=3 and gcd(10,15)=5. Therefore, it’s valid to color all elements the same color. Note that there are other colorings which satisfy Alice’s requirement in this test case.
In the second test case there is only one element of each color, so the coloring definitely satisfies Alice’s requirement.
只要发现小于sqrt(1000)的质数为10就好做了
#include<bits/stdc++.h>
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%lld",&a)
#define sd(a) scanf("%lf",&a)
#define sc(a) scahf("%c",&a);
#define ss(a) scanf("%s",a)
#define pi(a) printf("%d\n",a)
#define pl(a) printf("%lld\n",a)
#define pf(a) printf("%lf\n",a)
#define ms(a) memset(a,0,sizeof(a))
#define repi(i, a, b) for(register int i=a;i<=b;++i)
#define repd(i, a, b) for(register int i=a;i>=b;--i)
#define reps(s) for(register int i=head[s];i;i=Next[i])
#define ll long long
#define vi vector<int>
#define pii pair<int,int>
#define pll pair<long,long>
#define pil pair<int,long>
#define pli pair<long,int>
#define lowbit(x) ((x)&(-(x)))
#define ce(i, r) i==r?'\n':' '
#define pb push_back
#define fi first
#define se second
#define pr(x) cout<<#x<<": "<<x<<endl
using namespace std;
inline int qr() {
int f = 0, fu = 1;
char c = getchar();
while (c < '0' || c > '9') {
if (c == '-')fu = -1;
c = getchar();
}
while (c >= '0' && c <= '9') {
f = (f << 3) + (f << 1) + c - 48;
c = getchar();
}
return f * fu;
}
const int N = 1e3 + 10, M = 31;
int n, t, a[N];
bool v[M + 5], used[M + 5];
int num[M + 5], col[N];
vi pri;
inline void primes(int x) {
repi(i, 2, x) {
if (v[i])continue;
for (int j = i; j <= x / i; j++)v[i * j] = true;
}
repi(i, 2, x)if (!v[i])pri.pb(i);
}
int main() {
primes(M);
t = qr();
while (t--) {
n = qr();
repi(i, 1, n)a[i] = qr();
ms(used);
repi(i, 1, n)for (auto j:pri)
if (a[i] % j == 0) {
used[j] = true, col[a[i]] = j;
break;
}
int tot = 0;
repi(i, 0, pri.size() - 1)if (used[pri[i]])num[pri[i]] = ++tot;
pi(tot);
repi(i, 1, n)printf("%d%c", num[col[a[i]]], ce(i, n));
}
return 0;
}
