codeforces-873C. Strange Game On Matrix[模拟]

C. Strange Game On Matrix

Ivan is playing a strange game.

He has a matrix a with n rows and m columns. Each element of the matrix is equal to either 0 or 1. Rows and columns are 1-indexed. Ivan can replace any number of ones in this matrix with zeroes. After that, his score in the game will be calculated as follows:

1. Initially Ivan's score is 0;
2. In each column, Ivan will find the topmost 1 (that is, if the current column is j, then he will find minimum i such that a_i, j = 1). If there are no 1's in the column, this column is skipped;
3. Ivan will look at the next min(k, n - i + 1) elements in this column (starting from the element he found) and count the number of 1's among these elements. This number will be added to his score.

Of course, Ivan wants to maximize his score in this strange game. Also he doesn't want to change many elements, so he will replace the minimum possible number of ones with zeroes. Help him to determine the maximum possible score he can get and the minimum possible number of replacements required to achieve that score.

Input

The first line contains three integer numbers n, m and k (1 ≤ k ≤ n ≤ 100, 1 ≤ m ≤ 100).

Then n lines follow, i-th of them contains m integer numbers — the elements of i-th row of matrix a. Each number is either 0 or 1.

Output

Print two numbers: the maximum possible score Ivan can get and the minimum number of replacements required to get this score.

Examples

Input

4 3 2
0 1 0
1 0 1
0 1 0
1 1 1

Output

4 1

Input

3 2 1
1 0
0 1
0 0

Output

2 0

Note

In the first example Ivan will replace the element a_1, 2.

题意:

给出一个矩阵，可以把矩阵中的1变成0.每次找到每一列的第一个1，计算连续k个有多少1。要求最大的结果。

思路：直接模拟$sum$(此处检查Latex，请无视）

#include "bits/stdc++.h"
using namespace std;
const int maxn = 100 + 10;
int a[maxn][maxn];
int main(int argc, char const *argv[])
{
    int n,m,k;
    scanf("%d%d%d", &n, &m, &k);
    for(int i = 0; i < n; i++) {
        for(int j = 0;j < m; j++) {
            scanf("%d", &a[i][j]);
        }
    }
    int s1 = 0,s2 = 0;
    for(int j = 0;j < m; j++) {
        int p = 0;
        int maxx = 0;
        int t = 0;
        for(int i = 0;i < n; i++) {
            int cnt = 0;
            for(int l = i;l < n && l-i+1 <= k; l++) {
                if (a[l][j] == 1) cnt++;
            }
            if(maxx < cnt) {maxx = cnt; t = p;}
            p += a[i][j];
        }
        s1 += maxx;
        s2 += t;
    }
    printf("%d %d
", s1, s2);
    return 0;
}