分类目录归档：模拟

洛谷 P1022 计算器的改良

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#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
using namespace std;
int ratio = 0, constant = 0;
#define IfLetter if (ch <= 'Z'&&ch >= 'A' || ch <= 'z'&&ch >= 'a')
#define IfNumber if (ch >= '0'&&ch <= '9' || ch=='+' || ch=='-')
int main() {
    char letter;
    char ch;
    int num = 0;
    bool flag = true;
    while ((ch = getchar()) != '\n') {
        IfNumber{ // number(maybe with variable)
            ungetc(ch, stdin);
            int re = scanf("%d", &num); //specially prepared for the testcase form like "-a"
            if (re == 0) {
                getchar();
                ratio -= (flag ? 1 : -1);
            }
            ch = getchar();
            IfLetter{
                ratio += num * (flag ? 1 : -1);
                letter = ch;
            }
            else {
                constant += num * (flag ? 1 : -1);
                ungetc(ch, stdin);
            }
            num = 0;
        }
        else IfLetter{ //pure variable without ratio
            letter = ch;
            ratio+=(flag ? 1 : -1);
        }
        else if (ch == '=') {
            flag = false;
        }
    }
    printf("%c=%.3lf", letter, -1.0*constant / ratio);
}

洛谷 P1058 立体图

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虽然写的时间很长，但是AC之后随便画图可是很好玩的哦。。。

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#include<iostream>
#include<string>
#include<cstring>
#include<algorithm>
#include<functional>
#define ll long long
#define pii pair<int,int>
#define PINF 0x7fffffff
#define NINF 0x80000000
using namespace std;
int m, n, am, an;
int arr[55][55];
char output[500][500];
int calM(int m, int mh, int tm) { //每行挑最大的执行一次
    return (tm - m + 1) * 2 + max(1 + 3 * mh, 1 + 2 * (m - 1));
}
inline int calN(int m, int n) { //整体执行一次
    return 1 + 4 * n + 2 * m;
}
inline void printLine(int x, int y) {
    output[x][y] = output[x][y + 4] = '+';
    output[x][y + 1] = output[x][y + 2] = output[x][y + 3] = '-';
}
inline void printRow(int x, int y) {
    output[x][y] = output[x + 2][y - 2] = '+';
    output[x + 1][y - 1] = '/';
}
inline void printHeight(int x, int y) {
    output[x][y] = output[x + 3][y] = '+';
    output[x + 1][y] = output[x + 2][y] = '|';
}
inline void printSpace(int x, int y) {
    output[x + 1][y] = output[x + 1][y + 1] = output[x + 1][y + 2] = ' ';
    output[x + 3][y - 1] = output[x + 3][y] = output[x + 3][y + 1] = ' ';
    output[x + 4][y - 1] = output[x + 4][y] = output[x + 4][y + 1] = ' ';
    output[x + 2][y + 3] = output[x + 3][y + 3] = ' ';
}
inline void printCube(int x, int y) { //左上角坐标
    printLine(x, y);
    printLine(x + 2, y - 2);
    printLine(x + 5, y - 2);
    printRow(x, y);
    printRow(x, y + 4);
    printRow(x + 3, y + 4);
    printHeight(x + 2, y - 2);
    printHeight(x + 2, y + 2);
    printHeight(x, y + 4);
    printSpace(x, y);
}
inline void printCube2(int l, int r, int h) {
    printCube(am - 3 * h - 2 * (m-l+1), 1+2 * (m - l + 1) + (r - 1) * 4);
}
inline void printAns() {
    for (int i = 1; i <= am; i++) {
        cout << output[i] + 1 << endl;
    }
    cout << endl;
}
int main() {
    cin >> m >> n;
    an = calN(m, n);
    for (int i = 1; i <= m; i++) {
        int mmax = 0;
        for (int j = 1; j <= n; j++) {
            cin >> arr[i][j];
            mmax = max(mmax, arr[i][j]);
        }
        am = max(am, calM(i, mmax, m));
    }
    for (int i = 1; i <= am; i++) {
        for (int j = 1; j <= an; j++) {
            output[i][j] = '.';
        }
    }
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            for (int k = 1; k <= arr[i][j]; k++) {
                printCube2(i, j, k);
                //printAns();
            }
        }
    }
    printAns();
}

提示：把79行的注释符号去掉可以看到图形生成全过程哦；
写了个好玩的小程序，随机生成数组中的高度哦。。。
各位OIer可以随便试试。。。

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#include<iostream>
#include<string>
#include<cstring>
#include<algorithm>
#include<functional>
#include<ctime>
#define ll long long
#define pii pair<int,int>
#define PINF 0x7fffffff
#define NINF 0x80000000
using namespace std;
int m, n, am, an;
int arr[55][55];
char output[500][500];
int calM(int m, int mh, int tm) { //每行挑最大的执行一次
    return (tm - m + 1) * 2 + max(1 + 3 * mh, 1 + 2 * (m - 1));
}
inline int calN(int m, int n) { //整体执行一次
    return 1 + 4 * n + 2 * m;
}
inline void printLine(int x, int y) {
    output[x][y] = output[x][y + 4] = '+';
    output[x][y + 1] = output[x][y + 2] = output[x][y + 3] = '-';
}
inline void printRow(int x, int y) {
    output[x][y] = output[x + 2][y - 2] = '+';
    output[x + 1][y - 1] = '/';
}
inline void printHeight(int x, int y) {
    output[x][y] = output[x + 3][y] = '+';
    output[x + 1][y] = output[x + 2][y] = '|';
}
inline void printSpace(int x, int y) {
    output[x + 1][y] = output[x + 1][y + 1] = output[x + 1][y + 2] = ' ';
    output[x + 3][y - 1] = output[x + 3][y] = output[x + 3][y + 1] = ' ';
    output[x + 4][y - 1] = output[x + 4][y] = output[x + 4][y + 1] = ' ';
    output[x + 2][y + 3] = output[x + 3][y + 3] = ' ';
}
inline void printCube(int x, int y) { //左上角坐标
    printLine(x, y);
    printLine(x + 2, y - 2);
    printLine(x + 5, y - 2);
    printRow(x, y);
    printRow(x, y + 4);
    printRow(x + 3, y + 4);
    printHeight(x + 2, y - 2);
    printHeight(x + 2, y + 2);
    printHeight(x, y + 4);
    printSpace(x, y);
}
inline void printCube2(int l, int r, int h) {
    printCube(am - 3 * h - 2 * (m - l + 1), 1 + 2 * (m - l + 1) + (r - 1) * 4);
}
inline void printArray() {
    cout << "Printing each element of the array:" << endl;
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= m; j++) {
            cout << arr[i][j] << " ";
        }
        cout << endl;
    }
}
inline void printAns() {
    for (int i = 1; i <= am; i++) {
        cout << output[i] + 1 << endl;
    }
    cout << endl;
}
int main() {
    srand((unsigned)time(NULL));
    cout << "Please input the side length "n" of the cube array (n<=50):";
    int len;
    cin >> len;
    m = n = len;
    an = calN(m, n);
    cout << "Please input the max height "mh" of cubes in each array item (1<=mh<=100):";
    int mh;
    cin >> mh;
    cout << "Do you want to show the whole process of filling?(Answer with 0[No] or 1[Yes]):";
    bool flag;
    cin >> flag;
    for (int i = 1; i <= m; i++) {
        int mmax = 0;
        for (int j = 1; j <= n; j++) {
            arr[i][j] = rand() % mh + 1;
            mmax = max(mmax, arr[i][j]);
        }
        am = max(am, calM(i, mmax, m));
    }
    for (int i = 1; i <= am; i++) {
        for (int j = 1; j <= an; j++) {
            output[i][j] = '.';
        }
    }
    cout << "Automatically generated an array,printed in the following.Press ENTER to continue." << endl;
    printArray();
    cin.get(); cin.get();
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            for (int k = 1; k <= arr[i][j]; k++) {
                if (flag)cout << "Printing cube on postion(" << i << "," << j << "," << k << ")." << endl;
                printCube2(i, j, k);
                if (flag)printAns();
            }
        }
    }
    cout << "Printing Final Result:" << endl;
    printAns();
}

洛谷 谁拿了最多奖学金

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这题本身没啥好说的，非常简单，要说的是C++的一个特性。
这个题写的比较复杂，其实不用这么多空间，也不要排序，直接线性处理就行了。但是我写题写习惯了，又接着搞排序，就出了这么个问题。

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#include<bits/stdc++.h>
using namespace std;
int n;
struct stu{
    string name;
    int fs,cs;
    char sl,wp;
    int an;
    int money;
    stu(){
        money=0;
    }
    bool operator < (const stu& s2){
        return money>=s2.money;
    }
}stus[105];
int sum=0;
int main(){
    cin>>n;
    for(int i=1;i<=n;i++)
        cin>>stus[i].name>>stus[i].fs>>stus[i].cs>>stus[i].sl>>stus[i].wp>>stus[i].an;
    for(int i=1;i<=n;i++){
        if(stus[i].fs>80&&stus[i].an>=1)stus[i].money+=8000;
        if(stus[i].fs>85&&stus[i].cs>80)stus[i].money+=4000;
        if(stus[i].fs>90)stus[i].money+=2000;
        if(stus[i].fs>85&&stus[i].wp=='Y')stus[i].money+=1000;
        if(stus[i].cs>80&&stus[i].sl=='Y')stus[i].money+=850;
        sum+=stus[i].money;
    }
    sort(stus+1,stus+n+1);
    cout<<stus[1].name<<endl;
    cout<<stus[1].money<<endl;
    cout<<sum;
}

问题就出在比较的>=号上。查找了资料（地址http://blog.sina.com.cn/s/blog_532f6e8f01014c7y.html），我没细看，总结出来就是sort的cmp不能用带=号的。这个问题我最后就用stable_sort解决了。好用。代码如下：

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#include<iostream>
#include<algorithm>
using namespace std;
int n;
struct stu{
    string name;
    int fs,cs;
    char sl,wp;
    int an;
    int money;
    bool operator < (const stu& s2) const{
        return money>s2.money;
    }
}stus[205];
int sum1=0;
int main(){
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>stus[i].name>>stus[i].fs>>stus[i].cs>>stus[i].sl>>stus[i].wp>>stus[i].an;
    }
    for(int i=1;i<=n;i++){
        if(stus[i].fs>80&&stus[i].an>=1)stus[i].money+=8000;
        if(stus[i].fs>85&&stus[i].cs>80)stus[i].money+=4000;
        if(stus[i].fs>90)stus[i].money+=2000;
        if(stus[i].fs>85&&stus[i].wp=='Y')stus[i].money+=1000;
        if(stus[i].cs>80&&stus[i].sl=='Y')stus[i].money+=850;
        sum1+=stus[i].money;
    }
    stable_sort(stus+1,stus+n+1);
    cout<<stus[1].name<<endl;
    cout<<stus[1].money<<endl;
    cout<<sum1;
}

NOIP2017普及组前两道水题

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T1

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#include<iostream>
using namespace std;
int main() {
    double a, b, c;
    cin >> a >> b >> c;
    long long re = 0.2*a + 0.3*b + 0.5*c;
    cout << re;
}

T2

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#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
using namespace std;
bool cmp(string s1, string s2) {
    if (s1.length() != s2.length())return s1.length() < s2.length();
    return s1 < s2;
}
int main() {
    int n, q;
    vector<string> arr;
    cin >> n >> q;
    for (int i = 1; i <= n; i++) {
        string tmp;
        cin >> tmp;
        arr.push_back(tmp);
    }
    sort(arr.begin(), arr.end(), cmp);
    for (int i = 1; i <= q; i++) {
        int tmpi;
        string tmp;
        cin >> tmpi >> tmp;
        bool flag = false;
        for (vector<string>::iterator ptr = arr.begin(); ptr != arr.end(); ptr++) {
            if (ptr->length() - tmp.length() <=1e9 && ptr->substr(ptr->length() - tmp.length()) == tmp) {
                flag = true;
                cout << (*ptr) << endl;
                break;
            }
        }
        if (!flag)cout << "-1" << endl;
    }
}

洛谷 机器翻译

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题号：1540

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#include<bits/stdc++.h>
using namespace std;
int m, n;
int head = 0, tail = 0;
int arr[105];
int cnt;
int main(){
    ios::sync_with_stdio(false);
    cin >> m >> n;
    m++;
    for (int i = 0; i < m; i++){
        arr[i] = -1;
    }
    for (int i = 1; i <= n; i++){
        int word;
        cin >> word;
        bool flag = false;
        for (int i = head; i != tail; i = (i + 1) % m){
            if (arr[i] == word){
                flag = true;
                break;
            }
        }
        if (!flag){
            if (head == (tail + 1) % m)head = (head + 1) % m;
            cnt++;
            arr[tail] = word;
            tail = (tail + 1) % m;
        }
    }
    cout << cnt;
}

洛谷 序言页码 Preface Numbering

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也是一道恶心的题，相比上面那一道一点不会2333.
题号：1465

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char str[31][5] = {
    " ", "I", "II", "III", "IV", "V", "VI", "VII", "VIII",
    "IX", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX",
    "XC", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC",
    "CM", "M", "MM", "MMM"
};
char arr1[8] = "IVXLCDM";
int num[31] = {
    0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 200,
    300, 400, 500, 600, 700, 800, 900, 1000, 2000, 3000
};
int a[26];
void _find(int x){
    int j = 30;
    while (num[j] > x)j--;
    for (; j >= 1; j--){
        if (x >= num[j]){
            x -= num[j];
            for (int i = 0; i < strlen(str[j]); i++){
                a[str[j][i] - 'A']++;
            }
        }
        if (x == 0)return;
    }
}
int main(){
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++){
        _find(i);
    }
    if (a[int('I') - 65] != 0) printf("I %d\n", a[int('I') - 65]);
    if (a[int('V') - 65] != 0) printf("V %d\n", a[int('V') - 65]);
    if (a[int('X') - 65] != 0) printf("X %d\n", a[int('X') - 65]);
    if (a[int('L') - 65] != 0) printf("L %d\n", a[int('L') - 65]);
    if (a[int('C') - 65] != 0) printf("C %d\n", a[int('C') - 65]);
    if (a[int('D') - 65] != 0) printf("D %d\n", a[int('D') - 65]);
    if (a[int('M') - 65] != 0) printf("M %d\n", a[int('M') - 65]);
}

只有这样才能过……

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#include<cstdlib>
#include<cstring>
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
char str[31][5] = {
    " ", "I", "II", "III", "IV", "V", "VI", "VII", "VIII",
    "IX", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX",
    "XC", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC",
    "CM", "M", "MM", "MMM"
};
char arr[8] = "IVXLCDM";
int num[31] = {
    0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 200,
    300, 400, 500, 600, 700, 800, 900, 1000, 2000, 3000
};
int a[26];
void _find(int x){
    int j = 30;
    while (num[j] > x)j--;
    for (; j >= 1; j--){
        if (x >= num[j]){
            x -= num[j];
            for (int i = 0; i < strlen(str[j]); i++){
                a[str[j][i] - 'A']++;
            }
        }
        if (x == 0)return;
    }
}
int main(){
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++){
        _find(i);
    }
    for (int i = 0; i < 8; i++){
        if (a[arr[i] - 'A'] != 0)cout << (char)(arr[i]) << " " << a[arr[i] - 'A'] << endl;
    }
}

这样莫名其妙多出来一行输出@_500，根本不知道怎么回事……

洛谷 派对灯 Party Lamps

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USACO原题，比较恶心……看得题解
表示DFS只有三十分，就不用尝试DFS了，去看洛谷题解吧
题号：1468
https://www.luogu.org/wiki/show?name=%E9%A2%98%E8%A7%A3+P1468
作者: ☜闪耀星空☞ 更新时间: 2017-08-10 10:11

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#include<cstdlib>
#include<cstring>
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
int n, c, a, t;
bool light[10], dark[10],flag;
int map[10][10] = {
    { 0, 1, 1, 1, 1, 1, 1 }, //0
    { 0, 0, 0, 0, 0, 0, 0 }, //1
    { 0, 0, 1, 0, 1, 0, 1 }, //2
    { 0, 1, 0, 1, 0, 1, 0 }, //3
    { 0, 0, 1, 1, 0, 1, 1 }, //4
    { 0, 1, 0, 0, 1, 0, 0 }, //1,4
    { 0, 1, 1, 0, 0, 0, 1 }, //2,4
    { 0, 0, 0, 1, 1, 1, 0 }, //3,4
};
void print(int x){
    for (int i = 1; i <= 6; i++){
        if (light[i] && !map[x][i] || dark[i] && map[x][i])return;
    }
    flag = true;
    for (int i = 0; i < n; i++)cout << map[x][i % 6 + 1];
    cout << endl;
}
int main(){
    cin >> n >> c;
    while (cin >> t, t != -1)light[(t - 1) % 6 + 1] = true;
    while (cin >> t, t != -1)dark[(t - 1) % 6 + 1] = true;
    c = min(3, c);
    switch (c){
    case 0:print(0); break;
    case 1:print(1); print(2); print(4); print(3); break;
    case 2:print(1); print(7); print(2); print(4); print(3); print(6); print(0); break;
    case 3:print(1); print(7); print(2); print(4); print(5); print(3); print(6); print(0); break;
    }
    if (!flag)cout << "IMPOSSIBLE" << endl;
}

USACO Section 1.2 Transformations

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Transformations
A square pattern of size N x N (1 <= N <= 10) black and white square tiles is transformed into another square pattern. Write a program that will recognize the minimum transformation that has been applied to the original pattern given the following list of possible transformations: #1: 90 Degree Rotation: The pattern was rotated clockwise 90 degrees. #2: 180 Degree Rotation: The pattern was rotated clockwise 180 degrees. #3: 270 Degree Rotation: The pattern was rotated clockwise 270 degrees. #4: Reflection: The pattern was reflected horizontally (turned into a mirror image of itself by reflecting around a vertical line in the middle of the image). #5: Combination: The pattern was reflected horizontally and then subjected to one of the rotations (#1-#3). #6: No Change: The original pattern was not changed. #7: Invalid Transformation: The new pattern was not obtained by any of the above methods. In the case that more than one transform could have been used, choose the one with the minimum number above. PROGRAM NAME: transform INPUT FORMAT Line 1: A single integer, N Line 2..N+1: N lines of N characters (each either `@' or `-'); this is the square before transformation Line N+2..2*N+1: N lines of N characters (each either `@' or `-'); this is the square after transformation SAMPLE INPUT (file transform.in) 3 @[email protected] --- @@- @[email protected] @-- [email protected] OUTPUT FORMAT A single line containing the number from 1 through 7 (described above) that categorizes the transformation required to change from the `before' representation to the `after' representation. SAMPLE OUTPUT (file transform.out) 1 标准输入输出:

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#include<iostream>
#include<string>
#include<cstring>
int n;
namespace function {
    void reverse_str(char * s) {
        int len = strlen(s);
        for (int i = 0,j=len-1; i <j; i++,j--) {
            std::swap(s[i], s[j]);
        }
    }
    bool cmp(char arr1[11][11], char arr2[11][11]) {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (arr1[i][j] != arr2[i][j])return false;
            }
        }
        return true;
    }
    void rotate90(char arr1[11][11], char arr2[11][11]) {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                arr2[j][n-1-i] = arr1[i][j];
            }
        }
    }
    void rotate180(char arr1[11][11], char arr2[11][11]) {
        rotate90(arr1, arr2);
        rotate90(arr2, arr1);
        std::memcpy(arr2, arr1, sizeof(char)*11*11);
    }
    void rotate270(char arr1[11][11], char arr2[11][11]) {
        rotate90(arr1, arr2);
        rotate90(arr2, arr1);
        rotate90(arr1, arr2);
    }
    void reflection(char arr1[11][11], char arr2[11][11]) {
        std::memcpy(arr2, arr1, sizeof(char) * 11 * 11);
        for (int i = 0; i < n; i++) {
            reverse_str(arr2[i]);
        }
    }
    bool combination(char arr1[11][11], char arr2[11][11],char original[11][11],char transfered[11][11]) {
        std::memcpy(arr1, original, sizeof(char) * 11 * 11);
        reflection(arr1, arr2);
        std::memcpy(arr1, arr2, sizeof(char) * 11 * 11);
        rotate90(arr1, arr2);
        if (cmp(transfered, arr2))return true;
        rotate180(arr1, arr2);
        if (cmp(transfered, arr2))return true;
        std::memcpy(arr1, original, sizeof(char) * 11 * 11);
        reflection(arr1, arr2);
        std::memcpy(arr1, arr2, sizeof(char) * 11 * 11);
        rotate270(arr1, arr2);
        if (cmp(transfered, arr2))return true;
        return false;
    }
}
namespace std {
    char arr_original[11][11],arr_transfered[11][11];
    char temp1[11][11], temp2[11][11];
    int main() {
        function::rotate90(arr_original, temp1);
        if (function::cmp(temp1, arr_transfered))return 1;
        function::rotate90(temp1, temp2);
        if (function::cmp(temp2, arr_transfered))return 2;
        function::rotate90(temp2, temp1);
        if (function::cmp(temp1, arr_transfered))return 3;
        function::reflection(arr_original, temp1);
        if (function::cmp(temp1, arr_transfered))return 4;
        if(function::combination(temp1, temp2, arr_original,arr_transfered))return 5;
        if (function::cmp(arr_original, arr_transfered))return 6;
        return 7;
    }
}
int main() {
    std::ios::sync_with_stdio(false);
    std::cin >> n;
    for (int i = 0; i < n; i++) {
        std::cin >> std::arr_original[i];
    }
    for (int i = 0; i < n; i++) {
        std::cin >> std::arr_transfered[i];
    }
    std::cout<<std::main();
}

洛谷 [USACO1.1]坏掉的项链Broken Necklace

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题号：1203
并不知道为什么AC，本来45分瞎改了改AC的。。。这破题也是神烦……

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#include<iostream>
#include<algorithm>
#include<string>
#include<utility>
#include<queue>
#include<cstdlib>
#include<cstring>
#include<map>
using namespace std;
int n;
string s;
char search(int start, int ptr){
    for (int i = ptr; i - start +1 <= n; i++){
        if (s[i] != 'w')return s[i];
    }
    return 'w';
}
int run(int start,char c, char a){
    int tmax = 0;
    bool flag = false;
    int ptr = start;
    char cur;
    if (s[start] == a)return 0;
    while (1){
        if (!flag){
            if (s[ptr] == c || s[ptr] == 'w')tmax++;
            else {
                flag = true;
                tmax++;
                cur = search(start, ptr); //就这里，我觉得reference 2应该是ptr+1才对。。
            }
        }
        else{
            if (s[ptr] == cur || s[ptr] == 'w')tmax++;
            else break;
        }
        ptr++;
        if (ptr - start >= n){
           
            break;
        }
    }
    return tmax;
}
int main(){
    ios::sync_with_stdio(false);
    cin >> n >> s;
    s.append(s);
    int m = 0;
    for (int i = 0; i <= n; i++){
        int t = run(i, 'b', 'r');
        if (t>m)m = t;
        t = run(i, 'r', 'b');
        if (t > m)m = t;
    }
    cout << m;
}

计蒜客 NOIP模拟赛（一） D1T1

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题库链接：https://nanti.jisuanke.com/t/16445
本程序80分：

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#include<iostream>
using namespace std;
int prefixsum[2005][2005];
int n,k;
int safe_prefix(int x,int y){
    if(x>=1&&x<=n&&y>=1&&y<=n)return prefixsum[x][y];
    return 0;
}
int pong(int x,int y){
    int morse_cnt=0;
    bool flag=false;
    int cnt=0;
    for(int i=x-k+1;i<=x+k-1;i++){
        if(!flag&&cnt==k)flag=true;
        if(!flag)cnt++;
        else cnt--;
        if(i<1||i>n)continue;
        int jmin=y-cnt+1,jmax=y+cnt-1;
        morse_cnt+=safe_prefix(i,jmax)-safe_prefix(i,jmin-1);
    }
    return morse_cnt;
}
int main(){
    ios::sync_with_stdio(false);
    cin>>n>>k;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            int t;
            cin>>t;
            prefixsum[i][j]=prefixsum[i][j-1]+t;
        }
    }
    int m=0;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            int t=pong(i,j);
            if(t>m)m=t;
        }
    }
    cout<<m;
}