Solution
边(i,j)存在当且仅当[(ai,aj)≠1∨(ai+1,aj+1)≠1]。
就等价于求最大团。
随机乱搞好像这种题都过不了qwq
发现一个性质,ai与aj奇偶性相同时是必定有边的。考虑补图的最大独立集,显然这是一个二分图,跑最大匹配就好啦。
#include <bits/stdc++.h>
#define show(x) cerr << #x << " = " << x << endl
using namespace std;
typedef long long ll;
typedef pair<int, int> Pairs;
const int N = 505;
inline char get(void) {
static char buf[100000], *S = buf, *T = buf;
if (S == T) {
T = (S = buf) + fread(buf, 1, 100000, stdin);
if (S == T) return EOF;
}
return *S++;
}
template<typename T>
inline void read(T &x) {
static char c; x = 0; int sgn = 0;
for (c = get(); c < '0' || c > '9'; c = get()) if (c == '-') sgn = 1;
for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';
if (sgn) x = -x;
}
int n, ans, clc;
ll a[N];
int mp[N][N];
int usd[N], mat[N];
ll gcd(ll a, ll b) {
ll c = 1;
while (c) {
c = a % b;
a = b; b = c;
}
return a;
}
inline int match(int u) {
for (int i = 1; i <= n; i++)
if (usd[i] != clc && mp[u][i]) {
usd[i] = clc;
if (!mat[i] || match(mat[i])) {
mat[i] = u; return 1;
}
}
return 0;
}
int main(void) {
freopen("1.in", "r", stdin);
freopen("1.out", "w", stdout);
read(n); ans = n;
for (int i = 1; i <= n; i++) read(a[i]);
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
mp[i][j] = ((gcd(a[i], a[j]) == 1) && (gcd(a[i] + 1, a[j] + 1) == 1));
for (int i = 1; i <= n; i++)
if (a[i] & 1) {
++clc; ans -= match(i);
}
cout << ans << endl;
return 0;
}