Following website has the problem.
http://www.codechef.com/problems/PERMUT2
I have been trying to code solution for PERMUT2. My below solution is failing on some test cases. Kindly help me uncover flaws in the below code.
#include <stdio.h>
int a[100000];
int main()
{
int i, j, n, ret;
while(1)
{
scanf("%d", &n);
if(n == 0)
break;
ret = 0;
for(i = 0; i < n; i++)
scanf("%d", &a[i]);
for(i = 0; i < n; i++)
if(a[i] != i + 1)
ret++;
if(ret % 2 == 0)
printf("ambiguous\n");
else
printf("not ambiguous\n");
}
return 0;
}
You are not checking the right property. if(a[i] != i + 1) ret++; is not the right checking.
You want to check a[a[i] - 1] == i + 1 for all elements on the array:
bool ambiguous = true;
for(i = 0; i < n; i++) {
if (a[a[i] - 1] != i + 1) {
ambiguous = false;
break;
}
}
if(ambiguous)
printf("ambiguous\n");
else
printf("not ambiguous\n");
Related
I'm trying to write a simple simplex solver for linear optimization problems, but I'm having trouble getting it working. Every time I run it I get a vector subscript out of range (which is quite easy to find), but I think that its probably a core issue somewhere else in my impl.
Here is my simplex solver impl:
bool pivot(vector<vector<double>>& tableau, int row, int col) {
int n = tableau.size();
int m = tableau[0].size();
double pivot_element = tableau[row][col];
if (pivot_element == 0) return false;
for (int j = 0; j < m; j++) {
tableau[row][j] /= pivot_element;
}
for (int i = 0; i < n; i++) {
if (i != row) {
double ratio = tableau[i][col];
for (int j = 0; j < m; j++) {
tableau[i][j] -= ratio * tableau[row][j];
}
}
}
return true;
}
int simplex(vector<vector<double>>& tableau, vector<double>& basic, vector<double>& non_basic) {
int n = tableau.size() - 1;
int m = tableau[0].size() - 1;
while (true) {
int col = -1;
for (int j = 0; j < m; j++) {
if (tableau[n][j] > 0) {
col = j;
break;
}
}
if (col == -1) break;
int row = -1;
double min_ratio = numeric_limits<double>::infinity();
for (int i = 0; i < n; i++) {
if (tableau[i][col] > 0) {
double ratio = tableau[i][m] / tableau[i][col];
if (ratio < min_ratio) {
row = i;
min_ratio = ratio;
}
}
}
if (row == -1) return -1;
if (!pivot(tableau, row, col)) return -1;
double temp = basic[row];
basic[row] = non_basic[col];
non_basic[col] = temp;
}
return 1;
}
Problem: Count the number of ways to place K Bishops on N*N chess board.
https://onlinejudge.org/external/8/861.pdf
I came up with the below backtracking solution but it is not fast enough. I do not understand the DP solution recommended on other sites. Can you please help.
In the below code I am exhaustively searching for all solutions using a recursive backtracking approach.
#include <bits/stdc++.h>
using namespace std;
using Row = vector<bool>;
using Board = vector<Row>;
Board gBoard;
bool isOk(int n, int r, int c) {
for(int i = r-1, j = c-1; i >= 0 && j >= 0; --i, --j) {
if(gBoard[i][j]) {
return false;
}
}
for(int i = r+1, j = c+1; i < n && j < n; ++i, ++j) {
if(gBoard[i][j]) {
return false;
}
}
for(int i = r-1, j = c+1; i >= 0 && j < n; --i, ++j) {
if(gBoard[i][j]) {
return false;
}
}
for(int i = r+1, j = c-1; i < n && j >= 0; ++i, --j) {
if(gBoard[i][j]) {
return false;
}
}
return true;
}
void trace(const string &msg) {
cout << msg << "\n";
for(int i = 0; i < gBoard.size(); ++i) {
for(int j = 0; j < gBoard[i].size(); ++j) {
cout << gBoard[i][j] << " ";
}
cout << "\n";
}
cout << "\n";
}
void dfs(int n, int k, int r, int c, int cnt, int &sum) {
if(cnt == k) {
//trace("OK");
++sum;
return;
}
if(c >= n) {
c = 0;
r += 1;
}
int j = c;
for(int i = r; i < n; ++i) {
for( ; j < n; ++j) {
if(gBoard[i][j]) {
continue;
}
gBoard[i][j] = true;
if(isOk(n, i,j)) {
//trace("tmp");
dfs(n, k, i, j+1, cnt+1, sum);
}
gBoard[i][j] = false;
}
j = 0;
}
}
void solve(int n, int k) {
gBoard.assign(n, Row(n, false));
int sum = 0;
dfs(n, k, 0, 0, 0, sum);
cout << sum << "\n";
}
int main() {
while(true) {
int n,k;
cin >> n >> k;
if(!n && !k) {
break;
}
solve(n, k);
}
return 0;
}
Thank you
For some reason, whenever I run this code, it just opens; loads for a sec; then closes without doing anything. Whenever I try to narrow it down to a piece of code, it makes absolutely no sense, like the line int dirX.
#include <iostream>
#include <queue>
using namespace std;
void solve()
{
// ENTER CODE BELOW
struct Loc
{
int x, y;
Loc (int xx=0, int yy=0) : x(xx), y(yy) {}
};
int n=0, currX=1002, currY=1002, dx[]={-1,1,0,0},dy[]={0,0,-1,1}; string str=""; bool isFence[2010][2010]; queue<Loc> q;
int ret=-1;
for (int i = 0; i < 2005; i++) {
for (int j = 0; j < 2005; j++) {
isFence[i][j]=false;
}
}
cin >> n >> str;
isFence[currX][currY]=true;
int dirX, dirY;
for (auto i : str)
{
dirX=0; dirY=0;
if (i=='N') dirX=-1;
else if (i=='S') dirX=1;
else if (i=='W') dirY=-1;
else dirY=1;
for (int j = 0; j < 2; j++) {
currX += dirX;
currY += dirY;
isFence[currX][currY]=true;
}
}
Loc curr; int nx, ny;
for (int i = 0; i < 2005; i++)
{
for (int j = 0; j < 2005; j++)
{
cout << isFence[i][j] << endl;
if (isFence[i][j]) continue;
ret++;
q = std::queue<Loc>();
q.push(Loc(i,j));
isFence[i][j]=true;
while (!q.empty())
{
curr = q.front(); q.pop();
for (int k = 0; k < 4; k++) {
nx = curr.x+dx[k]; ny=curr.y+dy[k];
if (nx >= 0 && nx < 2005 && ny >= 0 && ny<2005 && !isFence[nx][ny]) {
isFence[nx][ny]=true;
q.push(Loc(nx, ny));
}
}
}
}
}
cout << ret;
// ENTER CODE ABOVE
}
int main()
{
solve();
}
Also, the reason I have all my code in the solve() function was because this is an assignment and I have to do it this way.
Sidenote: I wrote this code very quickly, so it's very badly formatted.
I am trying to use Malloc function to dynamically allocate memory but I also want to specify my data entry for operation rather than taking the user input.
I have found this code here which works fine, but I am working with a large data set and taking user input is not an option, so I want to keep using MALLOC but also define the data set.
like instead of following,
//Input Matrix1
for (i = 0; i < r1; i++)
for (j = 0; j < c1; j++)
scanf_s("%d", &mat1[i][j]);
I want something like
//mat1[2][2] = { {1,2},{2,3} }
to be inputed in the code
What would be the way to do it? I would really appreciate some advice. Thanks
#include<stdio.h>
#include<stdlib.h>
int main() {
int **mat1, **mat2, **res, i, j,k, r1, c1, r2, c2;
printf("\nEnter the Order of the First matrix...\n");
scanf_s("%d %d", &r1, &c1);
printf("\nEnter the Order of the Second matrix...\n");
scanf_s("%d %d", &r2, &c2);
if (c1 != r2) {
printf("Invalid Order of matrix");
exit(EXIT_SUCCESS);
}
mat1 = (int**)malloc(r1 * sizeof(int*));
for (i = 0; i < c1; i++)
mat1[i] = (int*)malloc(c1 * sizeof(int));
mat2 = (int**)malloc(r2 * sizeof(int*));
for (i = 0; i < c2; i++)
mat2[i] = (int*)malloc(c2 * sizeof(int));
res = (int**)calloc(r1, sizeof(int*));
for (i = 0; i < c2; i++)
res[i] = (int*)calloc(c2, sizeof(int));
/**/
//Input Matrix1
for (i = 0; i < r1; i++)
for (j = 0; j < c1; j++)
scanf_s("%d", &mat1[i][j]);
//Input Matrix2
for (i = 0; i < r2; i++)
for (j = 0; j < c2; j++)
scanf_s("%d", &mat2[i][j]);
//Printing Input Matrix 1 and 2
printf("\n Entered Matrix 1: \n");
for (i = 0; i < r1; i++) {
for (j = 0; j < c1; j++)
printf("%d ", mat1[i][j]);
printf("\n");
}
printf("\n Entered Matrix 2: \n");
for (i = 0; i < r2; i++) {
for (j = 0; j < c2; j++)
printf("%d ", mat2[i][j]);
printf("\n");
}
//int mat1[2][2] = { {1,2},{2,3} };
//int mat2[2][2] = { {1,3},{2,4} };
//Computation
//Multiplication
for (i = 0; i < r1; i++) {
for (j = 0; j < c2; j++) {
res[i][j] = 0;
for (k = 0; k < c1; k++)
res[i][j] += mat1[i][k] * mat2[k][j];
}
printf("\n");
}
printf("\nThe Multiplication of two matrix is\n");
for (i = 0; i < r1; i++) {
printf("\n");
for (j = 0; j < c2; j++)
printf("%d\t", res[i][j]);
}
printf("\n");
/* Addition
for(i=0;i<r1;i++)
for(j=0;j<c2;j++)
res[i][j]=mat1[i][j]+mat2[i][j];
printf("\nThe Addition of two matrix is\n");
for(i=0;i<r1;i++){
printf("\n");
for(j=0;j<c2;j++)
printf("%d\t",res[i][j]);
}
*/
return 0;
}
Please specify the format of your input data. Is it a csv file?
You can only specify data in the format int b[4] = {1, 2, 3, 4} when the size of b is fixed, i.e. known at compile time. But if all you matrices are known at compile time anyway, why bother doing dynamic allocation?
Also I cleaned up your code a bit:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define index(x, y, r) (x+r*y)
struct matrix {
int cols;
int rows;
double * data;
};
void print_mat(struct matrix * mat) {
int i, j;
for (i = 0; i < mat->rows; i++) {
for (j = 0; j < mat->cols; j++) {
printf("%f \t", mat->data[index(i, j, mat->rows)]);
}
printf("\n");
}
}
int mat_alloc(struct matrix *mat) {
mat->data = (double*)malloc(mat->rows*mat->cols*sizeof(double));
}
int read_mat(struct matrix *mat) {
int i, j;
for (i = 0; i < mat->rows; i++) {
for (j = 0; j < mat->cols; j++) {
scanf("%lf", &(mat->data[index(i, j, mat->rows)]));
}
}
}
int multiply(struct matrix * a, struct matrix * b, struct matrix * res) {
if(a->cols != b->rows){
printf("Matrix dimensions do not match!\n");
return 0;
}
res->rows = a->rows;
res->cols = b->cols;
mat_alloc(res);
memset(res->data, 0, res->cols*res->rows*sizeof(double));
int i, j, k;
for (i = 0; i < res->rows; i++) {
for (j = 0; j < res->cols; j++) {
for (k = 0; k < a->cols; k++) {
res->data[index(i, j, res->rows)] += a->data[index(i, k, a->rows)] * b->data[index(k, j, b->rows)];
}
}
}
}
int main() {
struct matrix mat1, mat2, res;
printf("\nEnter the Order of the First matrix...\n");
scanf("%d %d", &mat1.rows, &mat1.cols);
printf("\nEnter the Order of the Second matrix...\n");
scanf("%d %d", &mat2.rows, &mat2.cols);
mat_alloc(&mat1);
mat_alloc(&mat2);
read_mat(&mat1);
read_mat(&mat2);
printf("Scanned matrices: \n");
print_mat(&mat1);
printf("\n");
print_mat(&mat2);
printf("\n");
multiply(&mat1, &mat2, &res);
printf("Calculated result: \n");
print_mat(&res);
return 0;
}
How to implement a round robin schedule for an array of 4 elements [1,2,3,4]? The result of the algorithm should be able to display, for each element, the list of the players it will face in chronological order:
(1: 4,2,3)
(2: 3,1,4)
(3: 2,4,1)
(4: 1,3,2)
Line 1: 4,2,3 means that the player (1) will face in order the players (4), (2) and (3).
Of the same way, line 2: 3,1,4 indicates that the player (2) will face in order the players (3), (1) and (2).
We have implemented this code but we encounter a bug when we start filling in the name of the player. Do you have any idea about this problem?
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define NAME_MAX_LENGTH 20
#define NUM_MIN_PLAYERS 2
#define NUM_MAX_PLAYERS 20
enum Style
{
STYLE_COMPACT,
STYLE_TABLE
};
enum Format
{
FORMAT_ID,
FORMAT_NAME
};
struct PlayerList
{
unsigned int num_players;
char name[NUM_MAX_PLAYERS][NAME_MAX_LENGTH + 1];
};
struct Grid
{
unsigned int num_players;
unsigned int day[NUM_MAX_PLAYERS]
[NUM_MAX_PLAYERS];
};
void printList(struct PlayerList *list)
{
for (int i = 0; i < list->num_players; i++)
{
printf("%d:%s\n", i + 1, list->name[i]);
}
}
struct Grid calculer_berger(struct PlayerList *list)
{
struct Grid grid;
// algo pour remplir la grid
grid.num_players = list->num_players;
int i, j;
for (i = 0; i < list->num_players - 1; i++)
{
for (j = 0; j < list->num_players - 1; j++)
{
if (i == j)
{
/* edge cases */
grid.day[i][list->num_players - 1] = ((i + j) + (i + j) / list->num_players) % list->num_players;
grid.day[list->num_players - 1][j] = ((i + j) + (i + j) / list->num_players) % list->num_players;
grid.day[i][j] = 0;
}
else
{
grid.day[i][j] = ((i + j) + (i + j) / list->num_players) % list->num_players;
}
}
}
grid.day[0][list->num_players - 1] = list->num_players - 1;
grid.day[list->num_players - 1][list->num_players - 1] = 0;
grid.day[list->num_players - 1][0] = list->num_players - 1;
return grid;
}
void permuter(struct Grid *grid)
{
int tmp;
for (int i = 0; i < grid->num_players; i++)
{
for (int j = 1; j <= grid->num_players / 2; j++)
{
tmp = grid->day[i][j];
grid->day[i][j] = grid->day[i][grid->num_players - j];
grid->day[i][grid->num_players - j] = tmp;
}
}
}
void print_grid(struct Grid *grid, struct PlayerList *list)
{
for (int i = 0; i < grid->num_players; i++)
{
for (int j = 0; j < grid->num_players; j++)
{
if (j == 0)
{
printf("%d:", grid->day[i][j] + 1);
}
else
{
printf("%d", grid->day[i][j] + 1);
if (j < grid->num_players - 1)
{
printf(",");
}
}
}
printf("\n");
}
}
int main(int argc, char **argv)
{
struct PlayerList playerList;
char nom[NAME_MAX_LENGTH + 1];
int nbCharLu = 0;
while ((nbCharLu = fscanf(stdin, "%s", nom)) != -1)
{
strcpy(playerList.name[playerList.num_players], nom);
playerList.num_players++;
}
struct Grid myGrid = calculer_berger(&playerList);
printList(&playerList);
print_grid(&myGrid, &playerList);
printf("Apres la permut\n");
permuter(&myGrid);
print_grid(&myGrid, &playerList);
return 0;
}
Assuming you are storing the elements in an Integer array and that you would like to just display the results.
Here is one implementation....The code should accommodate "N" values because of the use of "sizeof"....
feel free to customize it further....
#include <stdio.h>
int main() {
int i,j;
int array[] = {1,2,3,4};
for(i = 0; i < sizeof(array)/sizeof(int);++i){
printf("(%d :",array[i]);
for(j = 0; j < sizeof(array)/sizeof(int);++j){
if(j == i)
continue;
printf("%d ",array[j]);
}
printf(")\n");
}
}
#include <stdio.h>
void main() {
int mid;
int num;
int j, temp;
int k = 0;
int num1;
int data[] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14};
num = sizeof(data)/sizeof(int);
mid = (sizeof(data)/sizeof(int))/2;
while(k < num - 1){
printf("Round %d ( ",k+1);
num1 = num;
for(int i = 0;i < mid;i++,num1--) /*pairing the competitors in each round*/
printf("%d:%d ",data[i],data[num1-1]);
for(int i = 0,j = num-1; i < num -2;i++,j--){ /* fixing the first competitor and rotating the others clockwise*/
temp = data[j];
data[j] = data[j-1];
data[j-1] = temp;
}
printf(")\n");
k++;
}
}