cf1043F. Make It One(dp 容斥原理)

题意

题目链接

给出\(n\)个数，问最少选几个数，使他们的\(gcd = 1\)

Sol

好神仙啊qwq。

首先，如果答案存在，那么最多为\(7\)(因为前\(7\)个质数乘起来\(>= 3e5\))

考虑dp，设\(f[i][j]\)表示选了\(i\)个数，他们\(gcd = j\)的方案数！

没错是方案数！

那么我们只要最后考虑一下\(f[i][1]\)是否有解就行了

设\(cnt[i]\)表示有多少个\(a_i\)存在\(i\)这个约数

转移的时候\(f[i][j] = C_{cnt[j]}^i - f[i][j * k], k >= 2\)

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<cmath>
#include<tr1/unordered_map> 
//#include<ext/pb_ds/assoc_container.hpp>
//#include<ext/pb_ds/hash_policy.hpp>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
//#define int long long
#define LL long long
#define ull unsigned long long
#define rg register
#define pt(x) printf("%d ", x);
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 22)], *p1 = buf, *p2 = buf;
//char obuf[1<<24], *O = obuf;
//void print(int x) {if(x > 9) print(x / 10); *O++ = x % 10 + '0';}
//#define OS  *O++ = ' ';
using namespace std;
//using namespace __gnu_pbds;
const int MAXN = 3e5 + 11, INF = 1e9 + 10, mod = 998244353, Mx = 3e5;
const double eps = 1e-9;
inline int read() {
    char c = getchar();
    int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, a[MAXN], f[12][MAXN], cnt[MAXN], fac[MAXN], ifac[MAXN];
int add(int x, int y) {
    if(x + y < 0) return x + y + mod;
    return x + y >= mod ? x + y - mod : x + y;
}
int mul(int x, int y) {
    return 1ll * x * y % mod;
}
int C(int N, int M) {
    if(N < M) return 0;
    return mul(fac[N], mul(ifac[M], ifac[N - M]));
}
int fp(int a, int p) {
    int base = 1;
    while(p) {
        if(p & 1) base = mul(base, a);
        a = mul(a, a); p >>= 1;
    }
    return base;
}
main() {
    N = read();
    for(int i = 1; i <= N; i++) {
        a[i] = read(), cnt[a[i]]++, f[1][a[i]]++;
        if(a[i] == 1) {puts("1"); return 0;}
    }
    fac[0] = 1; for(int i = 1; i <= N; i++) fac[i] = mul(i, fac[i - 1]);
    ifac[N] = fp(fac[N], mod - 2);
    for(int i = N; i >= 1; i--) ifac[i - 1] = mul(i, ifac[i]); 
    for(int i = 1; i <= Mx; i++) 
        for(int j = i + i; j <= Mx; j += i) cnt[i] += cnt[j];
    for(int i = 2; i <= 11; i++) {
        for(int j = Mx; j >= 1; j--) {
            f[i][j] = C(cnt[j], i);
            for(int k = j + j; k <= Mx; k += j) f[i][j] = add(f[i][j], -f[i][k]);
        }
        if(f[i][1] > 0) {printf("%d", i); return 0;}
    }
    puts("-1"); 
    return 0;
}