Codeforces 1043 F - Make It One

F - Make It One

思路:

dp + 容斥

首先, 答案不会超过7, 因为前7个质数的乘积大于3e5(最坏的情况是7个数, 每个数都缺少一个不同的因子)

所以从1到7依次考虑

dp[i][j]: 表示选取i个数且gcd==j的方案数

dp[i][j] = C(cntj, i) - ∑dp[i][k] (其中cntj表示ai中是j的倍数的个数, k表示所有j的倍数)

代码:

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int N = 3e5 + ;
const int MOD = 1e9 + ;
int a[N], cnt[N], mul_of[N];
int dp[][N];
int fac[N], invfac[N];
LL q_pow(LL n, LL k) {
    LL res = ;
    while(k) {
        if(k&) res = (res * n) % MOD;
        n = (n * n) % MOD;
        k >>= ;
    }
    return res;
}
void init() {
    fac[] = ;
    for (int i = ; i < N; i++) fac[i] = (1LL * fac[i-] * i) % MOD;
    invfac[N-] = q_pow(fac[N-], MOD-);
    for (int i = N-; i >= ; i--) invfac[i] = (1LL * invfac[i+] * (i+)) % MOD;
}
LL C(int n, int m) {
    if(n < m) return ;
    return (1LL * fac[n] * invfac[m]) % MOD * invfac[n-m] % MOD;
}
int main() {
    int n;
    init();
    scanf("%d", &n);
    for (int i = ; i <= n; i++) scanf("%d", &a[i]), cnt[a[i]]++;
    for (int i = ; i < N; i++) {
        for (int j = i; j < N; j += i) {
            mul_of[i] += cnt[j];
        }
    }
    for (int i = ; i <= ; i++) {
        for (int j = N-; j > ; j--) {
            int sum = ;
            for (int k = *j; k < N; k += j) sum = (sum + dp[i][k]) % MOD;
            dp[i][j] = (C(mul_of[j], i) - sum) % MOD;
        }
        if(dp[i][]) {
            printf("%d\n", i);
            return ;
        }
    }
    printf("-1\n");
    return ;
}