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 <bits/stdc++.h>
#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;
}