NOIP2023模拟1联测22 黑暗料理

NOIP2023模拟1联测22 黑暗料理

题目大意

自己看

思路

• 两个数相加能够产生质数的情况就是：1+1 或者 偶数+质数

  那么 $1$ 不能保留超过一个

• 建一个图，原点连向所有奇数点，所有偶数点连向汇点，奇数点和偶数点的和为奇数的就相连

  那么答案就是隔断原、汇两点的最小割。

• 判断质数用Miller_Rabin

code

#include 
#define fu(x , y , z) for(int x = y ; x <= z ; x ++)
#define fd(x , y , z) for(int x = y ; x >= z ; x --)
#define LL long long 
using namespace std;
const int N = 755;
int n , a[N * N * 2] , p[5] = {2 , 3 , 5 , 7 , 11} , cnt , ans , ans1 , hd[N * N * 2] , pos1[N] , pos2[N * N * 2] , cx1 , cx2 , cx[N] , cy[N] , vis[N];
struct E {
    int to , nt;
} e[N * N * 4];
void add (int x , int y) { e[++cnt].to = y , e[cnt].nt = hd[x] , hd[x] = cnt; }
bool cmp (int x , int y) { return a[x] > a[y]; }
LL ksm (LL x , LL y , LL mod) {
    if (!y) return 1;
    LL z = ksm (x , y / 2 , mod);
    z = z * z % mod;
    if (y & 1) z = z * x % mod;
    return z;
}
bool ck (LL x , LL y) {
    LL mod = x - 1 , xx;
    if (ksm (y , mod , x) != 1)
        return 0;
    do {
        xx = ksm (y , mod / 2 , x);
        if (xx != 1 && xx != x - 1)
            return 0;
        mod >>= 1;
    } while (mod % 2 == 0 && xx == 1);
    return 1;
}
bool mb (LL x) {
    if (x == 1)
        return 0;
    if (x == 2 || x == 3 || x == 5 || x == 7 || x == 11)
        return 1;
    fu (i , 0 , 4) {
        if (!ck (x , p[i])) 
            return 0;
    }
    return 1;
}
bool find (int x) {
    int y;
    for (int i = hd[x] ; i ; i = e[i].nt) {
        y = e[i].to;
        if (vis[y]) continue;
        vis[y] = 1;
        if (!cy[y] || find (cy[y])) {
            cx[x] = y , cy[y] = x;
            return 1;
        }
    }
    return 0;
}
int main () {
    freopen ("cooking.in" , "r" , stdin);
    freopen ("cooking.out" , "w" , stdout);
    int T , flg;
    scanf ("%d" , &T);
    while (T --) {
        scanf ("%d" , &n);
        fu (i , 1 , n)
            scanf ("%d" , &a[i]);
        cnt = ans1 = ans = flg = cx1 = cx2 = 0;
        fu (i , 1 , n * n * 2) hd[i] = cx[i] = cy[i] = 0;
        fu (i , 1 , n) {
            if (a[i] & 1) {
                if (a[i] == 1) {
                    if (!flg) flg = 1;
                    else {
                        ans1 ++;
                        continue;
                    }
                }
                pos1[++cx1] = i;
            }         
            else
                pos2[++cx2] = i;
        }
        fu (i , 1 , cx1) {
            fu (j , 1 , cx2) {
                if (a[pos1[i]] == 51 && a[pos2[j]] == 38) {
                    ans ++;
                    ans --;
                }
                if (mb (a[pos1[i]] + a[pos2[j]])) {
                    add (i , cx1 + j);
                    // cout << a[pos1[i]] << " " << a[pos2[j]] << "\n";
                }
            }
        }
        // exit (0);
        fu (i , 1 , cx1) {
            if (cx[i]) continue;
            fu (j , 1 , cx2) vis[j + cx1] = 0;
            ans += find (i);
        }
        // cout << ans;/
        printf ("%d\n" , n - ans1 - ans);
    }
    return 0;
}
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