Codeforces 840C On the Bench dp

On the Bench

两个数如果所有质因子的奇偶性相同则是同一个数，问题就变成了给你n个数， 相同数字不能相邻的方案数。

dp[ i ][ j ]表示前 i 种数字已经处理完， 还有 j 个位置需要隔开的方案数。

转移的话， 我们枚举第i + 1种数字分成的段数， 然后枚举有几段插到 j 个空格里面， 然后转移。

最后乘上各种数字个数的阶乘。

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 300 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}

int n, cnt, a[N], c[N];
int g[N][N], sum[N];
int dp[N][N];
ull hs[N];
map<int, int> Map;
vector<ull> oo;

int F[N], Finv[N], inv[N];

void init() {
    inv[1] = F[0] = Finv[0] = 1;
    for(int i = 2; i < N; i++) inv[i] = 1LL * (mod - mod / i) * inv[mod % i] % mod;
    for(int i = 1; i < N; i++) F[i] = 1LL * F[i - 1] * i % mod;
    for(int i = 1; i < N; i++) Finv[i] = 1LL * Finv[i - 1] * inv[i] % mod;
}
int comb(int n, int m) {
    if(n < 0 || n < m) return 0;
    return 1LL * F[n] * Finv[m] % mod * Finv[n - m] % mod;
}
int main() {
    init();
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        Map.clear();
        for(int j = 2; j * j <= a[i]; j++) {
            if(a[i] % j) continue;
            while(a[i] % j == 0) {
                Map[j]++;
                a[i] /= j;
            }
        }
        if(a[i] > 1) Map[a[i]]++;
        for(auto& t : Map) {
            if(t.se & 1) {
                hs[i] *= 23333;
                hs[i] += t.fi;
            }
        }
        oo.push_back(hs[i]);
    }
    sort(ALL(oo));
    oo.erase(unique(ALL(oo)), oo.end());
    for(int i = 1; i <= n; i++) a[i] = lower_bound(ALL(oo), hs[i]) - oo.begin() + 1;
    for(int i = 1; i <= n; i++) c[a[i]]++;
    cnt = n;
    n = SZ(oo);
    for(int i = 1; i <= n; i++) sum[i] = sum[i - 1] + c[i];
    g[0][0] = 1;
    for(int i = 0; i <= cnt; i++) {
        for(int j = 0; j <= cnt; j++) {
            if(!g[i][j]) continue;
            for(int k = 1; i + k <= cnt; k++) {
                add(g[i + k][j + 1], g[i][j]);
            }
        }
    }
    dp[1][c[1] - 1] = 1;
    for(int i = 1; i < n; i++) {
        for(int j = 0; j <= cnt; j++) {
            if(!dp[i][j]) continue;
            int num = c[i + 1];
            for(int k = 1; k <= num && k <= sum[i] + 1; k++) {
                int way = g[num][k];
                for(int z = max(0, k - sum[i] - 1 + j); z <= k; z++) {
                    add(dp[i + 1][j - z + num - k], 1LL * way * comb(j, z) % mod * comb(sum[i] + 1 - j, k - z) % mod * dp[i][j] % mod);
                }
            }
        }
    }
    int ans = dp[n][0];
    for(int i = 1; i <= n; i++)
        ans = 1LL * ans * F[c[i]] % mod;
    printf("%d\n", ans);
    return 0;
}

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