Codeforces Round #166 (Div. 2) B. Prime Matrix（素数筛选，简单）

You've got an n × m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.

You are really curious about prime numbers. Let us remind you that a prime number is a positive integer that has exactly two distinct positive integer divisors: itself and number one. For example, numbers 2, 3, 5 are prime and numbers 1, 4, 6 are not.

A matrix is prime if at least one of the two following conditions fulfills:

the matrix has a row with prime numbers only;
the matrix has a column with prime numbers only;

Your task is to count the minimum number of moves needed to get a prime matrix from the one you've got.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 500) — the number of rows and columns in the matrix, correspondingly.

Each of the following n lines contains m integers — the initial matrix. All matrix elements are positive integers. All numbers in the initial matrix do not exceed 10⁵.

The numbers in the lines are separated by single spaces.

Output

Print a single integer — the minimum number of moves needed to get a prime matrix from the one you've got. If you've got a prime matrix, print 0.

Sample test(s)

Input

Output

Input

2 3
4 8 8
9 2 9

Output

Input

2 2
1 3
4 2

Output

Note

In the first sample you need to increase number 1 in cell (1, 1). Thus, the first row will consist of prime numbers: 2, 2, 3.

In the second sample you need to increase number 8 in cell (1, 2) three times. Thus, the second column will consist of prime numbers: 11, 2.

In the third sample you don't have to do anything as the second column already consists of prime numbers: 3, 2.

求出每个元素变成素数所需的操作次数a[i][j]，然后分别求出每一行、每一列的∑a[i][j]，取最小值即可。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 
 5 using namespace std;
 6 
 7 const int size = 502;
 8 int mat[size][size], n, m;
 9 bool p[100100] = {0, 1, 0, 0};  //素数为0
10 
11 void Prime()  //筛选素数
12 {
13     int a = sqrt(100105);
14     for(int i = 2; i < a; i++)
15     {
16         if(!p[i])
17             for(int j = 2; j * i < 100100; j++)
18                 p[i*j] = 1;
19     }
20    // for(int i = 1; i < 100; i++) cout << i << ": " << p[i] << endl;
21 }
22 
23 int Find(int i)  //寻找比合数i大且与i最接近的素数，返回两者差值
24 {
25     int j = i + 1;
26     for(; p[j]; j++){}
27     return j - i;
28 }
29 
30 int main()
31 {
32     Prime();
33     while(scanf("%d %d", &n, &m) != EOF)
34     {
35         int e, cmin = 100000000, rmin = 100000000;
36         for(int i = 0; i < n; i++)
37         {
38             for(int j = 0; j < m; j++)
39             {
40                 scanf("%d", &e);
41                 if(p[e]) mat[i][j] = Find(e);  //e为合数
42                 else mat[i][j] = 0;
43             }
44         }
45         //找出列中所需操作的最大值
46         for(int j = 0; j < m; j++)
47         {
48             int sum = 0;
49             for(int i = 0; i < n; i++)
50                 sum += mat[i][j];
51             if(cmin > sum) cmin = sum;
52         }
53         //找出列中所需操作的最大值
54         for(int i = 0; i < n; i++)
55         {
56             int sum = 0;
57             for(int j = 0; j < m; j++)
58                 sum += mat[i][j];
59             if(rmin > sum) rmin = sum;
60         }
61         if(rmin > cmin) printf("%d\n", cmin);
62         else printf("%d\n", rmin);
63     }
64     return 0;
65 }

2013-02-17 21:47:02