D. Perfect Groups+唯一分解+组合

最新推荐文章于 2024-07-31 21:27:29 发布

ujn20161222

最新推荐文章于 2024-07-31 21:27:29 发布

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本文链接：https://blog.youkuaiyun.com/ujn20161222/article/details/80314085

本文介绍了一种算法，用于解决给定整数数组中寻找连续子数组的问题，这些子数组可以被划分成最少的组，使得每组内任意两个整数的乘积为完全平方数。文中提供了一个C++实现示例，包括输入输出处理、数据预处理及核心算法。

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SaMer has written the greatest test case of all time for one of his problems. For a given array of integers, the problem asks to find the minimum number of groups the array can be divided into, such that the product of any pair of integers in the same group is a perfect square.

Each integer must be in exactly one group. However, integers in a group do not necessarily have to be contiguous in the array.

SaMer wishes to create more cases from the test case he already has. His test case has an array $A$

n

integers, and he needs to find the number of contiguous subarrays of

A

that have an answer to the problem equal to

k

for each integer

k

between

1

and

n

(inclusive).

Input

The first line of input contains a single integer $n$

(

1 \leq n \leq 5000

), the size of the array.

The second line contains $n$

integers

a_{1}

a_{2}

\dots

a_{n}

(

- 10^{8} \leq a_{i} \leq 10^{8}

), the values of the array.

Output

Output $n$

space-separated integers, the

k

-th integer should be the number of contiguous subarrays of

A

that have an answer to the problem equal to

k

Examples

Input

Copy

2
5 5

Output

Copy

3 0

Input

Copy

5
5 -4 2 1 8

Output

Copy

5 5 3 2 0

Input

Copy

1
0

Output

Copy

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define all(a) (a).begin(),(a).end()
#define pll pair<ll,ll>
#define vi vector<int>
#define pb push_back
ll rd(){
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

const int P=1e5;
const int N=5000+10;
bool np[P];
vi pset;
void build(){
    rep(i,2,P-1)if(not np[i]){
        pset.push_back(i);
        for(int j=i;j<P;j+=i)np[j]=true;
    }
}
int __=1;
int n,a[N],aa[N];

int mask(int val){
    if(val==0)return val;
    int ret=val;
    val=abs(val);
    for(auto i:pset){
        if((ll)i*i>val)break;
        while(val%(i*i)==0){
            val/=(i*i);
            ret/=(i*i);
        }
    }
    return ret;
}



void init(){
    n=rd();
    vi li;
    rep(i,0,n-1){
        a[i]=mask(rd());
        li.push_back(a[i]);
    }
    sort(all(li));
    li.resize(unique(li.begin(),li.end())-li.begin());
    rep(i,0,n-1)
    aa[i]=lower_bound(all(li),a[i])-li.begin();
}

int cnt[N],ans[N],ac;

void solve(){
    rep(i,0,n){
        ac=0;
         rep(j,0,n-1)cnt[j]=0;
         rep(j,i,n-1){
             if(a[j]){
                 cnt[aa[j]]++;
                 if(cnt[aa[j]]==1)
                    ac++;
             }
             ans[max(1,ac)]++;
         }
    }
    rep(i,1,n)
    printf("%d ",ans[i]);
    puts("");
}

int main(){
   //freopen("in.txt","r",stdin);
   build();
   while(__--){
    init();
    solve();
   }
}