二分查找/暴力 Codeforces Round #166 (Div. 2) B. Prime Matrix

题目传送门

 /*
     二分查找/暴力：先埃氏筛选预处理，然后暴力对于每一行每一列的不是素数的二分查找最近的素数，更新最小值
 */
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 using namespace std;

 const int MAXN = 5e2 + ;
 const int MAXM = 1e6 + ;
 const int INF = 0x3f3f3f3f;
 int a[MAXN][MAXN];
 int mn_r[MAXN];
 int mn_c[MAXN];
 bool is_prime[MAXM];
 int prime[MAXM];
 int n, m, x, y;
 int tot;

 void solve(void)
 {
     for (int i=; i<=1e6; ++i)    is_prime[i] = true;
     is_prime[] = false;
     for (int i=; i<=1e6; ++i)
     {
         if (is_prime[i])
         {
             prime[++tot] = i;
             for (int j=i*; j<=1e6; j+=i)    is_prime[j] = false;
         }
     }
 }

 bool check_r(void)
 {
     memset (mn_r, , sizeof (mn_r));
     bool ok = true;
     for (int i=; i<=n; ++i)
     {
         for (int j=; j<=m; ++j)
         {
             if (!is_prime[a[i][j]])
             {
                 ok = false;    mn_r[j]++;
             }
         }
     }

     return ok;
 }

 bool check_c(void)
 {
     memset (mn_c, , sizeof (mn_c));
     bool ok = true;
     for (int j=; j<=m; ++j)
     {
         for (int i=; i<=n; ++i)
         {
             if (!is_prime[a[i][j]])
             {
                 ok = false;    mn_c[j]++;
             }
         }
     }

     return ok;
 }

 int sum_r(void)
 {
     int ans = INF;    int tmp = ;
     for (int i=; i<=n; ++i)
     {
         tmp = ;    bool ok = true;
         for (int j=; j<=m; ++j)
         {
             if (!is_prime[a[i][j]])
             {
                 int p = lower_bound (prime+, prime++tot, a[i][j]) - prime;
                 if (p <= tot)    tmp += prime[p] - a[i][j];
                 else    {ok = false;    break;}
             }
         }
         if (ok) ans = min (ans, tmp);
     }

     return ans;
 }

 int sum_c(void)
 {
     int ans = INF;    int tmp = ;
     for (int j=; j<=m; ++j)
     {
         tmp = ;    bool ok = true;
         for (int i=; i<=n; ++i)
         {
             if (!is_prime[a[i][j]])
             {
                 int p = lower_bound (prime+, prime++tot, a[i][j]) - prime;
                 if (p <= tot)    tmp += prime[p] - a[i][j];
                 else    {ok = false;    break;}
             }
         }
         if (ok) ans = min (ans, tmp);
     }

     return ans;
 }

 int main(void)        //Codeforces Round #166 (Div. 2) B. Prime Matrix
 {
 //    freopen ("B.in", "r", stdin);

     solve ();
     while (scanf ("%d%d", &n, &m) == )
     {
         for (int i=; i<=n; ++i)
         {
             for (int j=; j<=m; ++j)
                 scanf ("%d", &a[i][j]);
         }

         if (check_r () || check_c ())    {puts ("");    continue;}
         else    printf ("%d\n", min (sum_r (), sum_c ()));
     }

     return ;
 }