The problem is this http://www.spoj.com/problems/CMPLS/
I thought of implementing newtons divided difference and find out the polynomial. But I am not able to get the correct answer. Some basic test cases are passing but some are not.
# include <stdio.h>
int main()
{
int i, j, k, t, m, x[100][100], s, c, n;
long ans, mul;
scanf ("%d", &t);
while (t--) {
scanf ("%d%d", &s, &c);
n = s;
for (i = 0; i < n; i++)
scanf ("%d", &x[i][0]);
for (j = 1; j < n; j++)
for (i = 0; i < n; i++) {
x[i][j] = x[i+1][j-1] - x[i][j-1];
}
k = n;
printf("\n");
for (i = 0; i < n; i++) {
for(j = 0; j < k; j++)
printf ("%d ", x[i][j]);
printf ("\n");
k--;
}
printf ("\n\n");
for (m = 1; m <= c; m++) {
ans = x[0][0];
for ( i = 1; i < n - 1; i++) {
mul = x[0][i];
for (j = 1; j <= i; j++)
mul = mul * (s + m - j);
ans = ans + mul;
}
printf ("%ld ", ans);
}
printf ("\n");
}
return 0;
}
Related
here below i have given multiplication of matrices in c language using for loop but can any help me make a more simplified version or can any help me make it using while loop
i want a simplified version
i want a code in while loop
:) just learning
#include <stdio.h>
#include <stdlib.h>
int main()
{
int a[10][10], b[10][10], mul[10][10], r, c, i, j, k;
system("cls");
printf("enter the number of row=");
scanf("%d", &r);
printf("enter the number of column=");
scanf("%d", &c);
printf("enter the first matrix element=\n");
for (i = 0; i < r; i++)
{
for (j = 0; j < c; j++)
{
scanf("%d", &a[i][j]);
}
}
printf("enter the second matrix element=\n");
for (i = 0; i < r; i++)
{
for (j = 0; j < c; j++)
{
scanf("%d", &b[i][j]);
}
}
printf("multiply of the matrix=\n");
for (i = 0; i < r; i++)
{
for (j = 0; j < c; j++)
{
mul[i][j] = 0;
for (k = 0; k < c; k++)
{
mul[i][j] += a[i][k] * b[k][j];
}
}
}
//for printing result
for (i = 0; i < r; i++)
{
for (j = 0; j < c; j++)
{
printf("%d\t", mul[i][j]);
}
printf("\n");
}
return 0;
}
It will not make code simpler only harder to read.
One of the loops example:
printf("multiply of the matrix=\n");
i = 0;
while (i < r)
{
j = 0;
while(j < c)
{
mul[i][j] = 0;
k = 0;
while(k < c)
{
mul[i][j] += a[i][k] * b[k][j];
k++;
}
j++;
}
i++;
}
I need to multiply two square matrixes A and B 15x15.
Unfortunately, I'm getting this kind of error.
I know the problem is in pointers while calculating matrix C.
C[i][j] += *(A + k) * *(B + k)
I hope you can explain me what's wrong. I'm a beginner xD.
Thank you in advance.
#include <stdio.h>
#define N 15
#define _CRT_SECURE_NO_WARNINGS
int main() {
int A[N][N];
int B[N][N];
int C[N][N];
printf("Input matrix A.\n");
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
printf("Enter your element:\n");
scanf_s("%d", &A[i][j]);
}
printf("\n");
}
printf("Input matrix B.\n");
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
printf("Enter your element:\n");
scanf_s("%d", &B[i][j]);
}
printf("\n");
}
printf("Matrix A.\n");
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
printf("%d\t", A[i][j]);
}
printf("\n");
}
printf("Matrix B.\n");
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
printf("%d\t", B[i][j]);
}
printf("\n");
}
for (int i = 0; i < 15; i++) {
for (int j = 0; j < 15; j++) {
C[i][j] = 0;
for (int k = 0; k < 14; k++) {
C[i][j] += *(A + k) * *(B + k);
k++;
}
}
}
printf("Your result:\n");
printf("Matrix C.\n");
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
printf("%d\t", C[i][j]);
}
printf("\n");
}
return 0;
}
The problem in the multiplication is that A+k and B+k have type int (*)[15] which means dereferencing it once only makes a pointer out of them; furthermore, you need to take row and column items individually, which means A[i][k] and B[k][j], right? (also, there's no point on using confusing syntax, as the underlying operation is exactly the same).
Here's a fixed and improved version:
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#define N 15
/* Improvement 1 (type abstraction) */
typedef int NxN_int_matrix[N][N];
/* Improvement 2 (input function & wrapper) */
#define input_matrix(var) input_matrix_ex((var), #var)
static void input_matrix_ex(NxN_int_matrix dst, char *name)
{
printf("Input matrix %s.\n", name);
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
/* Improvement 3 (nicer prompt) */
printf("%s[%2d][%2d]: ", name, i, j);
fflush(stdout);
scanf_s("%d", &dst[i][j]);
}
}
printf("\n");
}
/* Improvement 4 (print function) */
#define print_matrix(var) print_matrix_ex(#var, (var))
static void print_matrix_ex(char *name, NxN_int_matrix M)
{
printf("Matrix %s.\n", name);
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
printf("%d\t", M[i][j]);
}
printf("\n");
}
}
/* Improvement 5 (move multiplication to a function too, and fix it) */
static void mult_matrix(NxN_int_matrix dst, NxN_int_matrix a, NxN_int_matrix b)
{
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
/* Improvement 6 (don't write out intermediate values) */
int tmp = 0;
for (int k = 0; k < N; k++)
tmp += a[i][k] * b[k][j];
dst[i][j] = tmp;
}
}
}
int main()
{
NxN_int_matrix A, B, C;
input_matrix(A);
input_matrix(B);
print_matrix(A);
print_matrix(B);
mult_matrix(C, A, B);
printf("Your result:\n");
print_matrix(C);
return 0;
}
/* Possible further improvements:
* - using a transposed B might make multiplication faster
*/
Newb Here!
Trying some things with C Language,
I tried to print Half left triangle/pyramid and Half right triangle/pyramid together in one print.
heres the code!
#include <stdio.h>
#include <conio.h>
int main()
{
int i, j, rows, k;
printf (" Enter a number to define the rows: \n ");
scanf("%d", &rows);
printf("\n");
for (i = 1; i <= rows; ++i) // outer loop
{
for (j = 1; j <= i; ++j) // inner loop
{
printf ("*"); // print the Star
}
printf ("\n");
}
{
for (i = 1; i <= rows; i++)
{
for (j = i; j < rows; j++)
{
printf(" ");
}
for (k = 1; k <= i; k++)
{
printf("*"); // print the Star
}
printf ("\n");
}
}
}
#include <stdio.h>
#include <conio.h>
int main()
{
int i, j, rows, k;
int hollow_spacing = 2;
int left_padding = 10;
printf (" Enter a number to define the rows: \n ");
scanf("%d", &rows);
printf("\n");
for (i = 0; i < rows; i++)
{
int curr_left_pad = left_padding + rows - i;
while(curr_left_pad-- > 0){
printf (" ");
}
for (int j = i; j >= 0; j--)
printf ("*");
for (int j = 0; j < hollow_spacing; ++j)
printf (" ");
for (int j = i; j >= 0; j--)
printf ("*");
printf ("\n");
}
}
My code never stops asking for input so I think I must've made an infinite loop, but I can't find where the error is. I've also noticed that when inserting the input line by line, it prints the result of one loop after inserting the first line of the second loop, which seems incorrect to me. Please help me debug. (For further context, the code is supposed to receive a number we'll call n, and then scan 3n more lines, which are basically n bundles of 3 similar lines. The 2nd and 3rd lines are two words with the same num of characters, and the 1st line is that num. The output is whether or not these words are anagrams.)
#include <stdio.h>
int main() {
int n, l;
scanf("%d\n", &n);
for (int i = 1; i <= n; i++) {
scanf("%d\n", &l);
char A[l], B[l];
for (int j = 0; j < l; j++) {
scanf("%c", &A[j]);
scanf("\n");
}
for (int j = 0; j < l; j++) {
scanf("%c", &B[j]);
scanf("\n");
}
for (int k = 0; k < l; k++) {
int result = 0;
for (int j = 0; j < l; j++) {
if (A[k] == B[j]) {
result = 1;
}
}
if (!result) {
printf("\nNO\n");
return 0;
}
}
printf("\nYES\n");
}
}
Example:
input:
2
6
listen
silent
4
Evil
live
output:
YES
NO
It's because you are asking for input over and over and over again,
Not sure what you are trying to achieve, but start by removing the extra scanfs
try this,
#include <stdio.h>
int main() {
int n, l;
scanf("%d\n", &n);
for (int i = 1; i <= n; i++) {
char A[l], B[l];
for (int k = 0; k < l; k++) {
int result = 0;
for (int j = 0; j < l; j++) {
if (A[k] == B[j]) {
result = 1;
}
}
if (!result) {
printf("\nNO\n");
return 0;
}
}
printf("\nYES\n");
}
}
Your scanf calls wait for an extra \n which requires more input to enter.
To fix this, remove the \n from your scanf calls. Also remove the extra calls when you enter A and B:
I have added some debug code to demonstrate where you are in your program execution while you enter your input.
#include <stdio.h>
int main() {
int n, l;
int res;
res=scanf("%d", &n);
printf ("res=%d, n=%d\n", res, n);
for (int i = 1; i <= n; i++) {
res = scanf("%d", &l);
printf ("res=%d, l=%d\n", res, l);
// char A[l], B[l];
char A[l+1], B[l+1];
for (int j = 0; j < l; j++) {
scanf(" %c", &A[j]);
}
printf ("A done\n");
for (int j = 0; j < l; j++) {
scanf(" %c", &B[j]);
}
printf ("B done\n");
A[l] = 0; B[l] = 0;
printf ("A=\"%s\" - B=\"%s\"\n", A, B);
...
Now your input should work properly.
But your code also contains another error.
You will treat "12344" and "11234" as correct match wchich is wrong.
To fix this you need to remove each matching character from B:
for (int k = 0; k < l; k++) {
int result = 0;
// We compare A[k] with the remaining characters in B only
for (int j = k; j < l; j++) {
if (A[k] == B[j]) {
result = 1;
B[j] = B[k]; // Replace matching character with non-matching character we checked earlier.
break;
}
}
if (!result) {
printf("\nNO\n");
return 0;
}
}
printf("\nYES\n");
This code stops as soon as the first match is found and removed that character from B.
The entries in B are rearranged to keep the unused entries at the end.
The full code looks like this:
#include <stdio.h>
int main() {
int n, l;
int res;
res=scanf("%d", &n);
printf ("res=%d, n=%d\n", res, n);
for (int i = 1; i <= n; i++) {
res = scanf("%d", &l);
printf ("res=%d, l=%d\n", res, l);
// char A[l], B[l];
char A[l+1], B[l+1];
for (int j = 0; j < l; j++) {
scanf(" %c", &A[j]);
}
printf ("A done\n");
for (int j = 0; j < l; j++) {
scanf(" %c", &B[j]);
}
printf ("B done\n");
A[l] = 0; B[l] = 0;
printf ("A=\"%s\" - B=\"%s\"\n", A, B);
for (int k = 0; k < l; k++) {
int result = 0;
for (int j = k; j < l; j++) {
if (A[k] == B[j]) {
result = 1;
B[j] = B[k];
break;
}
}
if (!result) {
printf("\nNO\n");
return 0;
}
}
printf("\nYES\n");
}
}
output:
~/stackoverflow$ ./test
4
res=1, n=4
2
res=1, l=2
12
A done
21
B done
A="12" - B="21"
YES
6
res=1, l=6
liver1
A done
evil1r
B done
A="liver1" - B="evil1r"
YES
3
res=1, l=3
111
A done
111
B done
A="111" - B="111"
YES
4
res=1, l=4
abcd
A done
dcbb
B done
A="abcd" - B="dcbb"
NO
I'm trying to write a programm that solves system of equations Ax=B using Gauss-Jacobi iteration method.
#include <math.h>
#include <stdlib.h>
#include <stdio.h>
int main(void) {
double **a, *b, *x, *f, eps = 1.e-2, c;
int n = 3, m = 3, i, j, bool = 1, d = 3;
/* printf("n=") ; scanf("%d", &n);
printf("m=") ; scanf("%d", &n) */
a =malloc(n * sizeof *a);
for (i = 0; i < n; i++)
a[i] = (double*)malloc(m * sizeof(double));
b = malloc(m * sizeof *b);
x = malloc(m * sizeof *x) ;
f = malloc(m * sizeof *f) ;
for (i = 0; i < n; i++) {
for (j = 0; j < m; j++) {
printf("a[%d][%d]=", i, j);
scanf("%le", &a[i][j]);
if(fabs(a[i][i])<1.e-10) return 0 ;
}
printf("\n") ;
}
printf("\n") ;
for (i = 0; i < n; i++) {
for (j = 0; j < m; j++) {
printf("a[%d][%d]=%le ", i, j, a[i][j]);
}
printf("\n") ;
}
for (j = 0; j < m; j++) {
printf("x[%d]=", j);
scanf("%le", &x[j]);
} //intial guess
printf("\n") ;
for (j = 0; j < m; j++) {
printf("b[%d]=", j);
scanf("%le", &b[j]);
}
printf("\n") ;
while (1) {
bool = 0;
for (i = 0; i < n; i++) {
c = 0.0;
for (j = 0; j < m; j++)
if (j != i)
c += a[i][j] * x[j];
f[i] = (b[i] - c) / a[i][i];
}
for (i = 0; i < m; i++)
if (fabs(f[i] - x[i]) > eps)
bool = 1;
if (bool == 1)
for (i = 0; i < m; i++)
x[i] = f[i];
else if (bool == 0)
break;
}
for (j = 0; j < m; j++)
printf("%le\n", f[j]);
return 0;
}
The condition of stoping the loop is that previous approximation minus current approximation for all x is less than epsilon.
It seems like i did everything according to algorithm,but the programm doesn't work.
Where did i make a mistake?
While not the most strict condition, the usual condition requiered to guarantee convergence in the Jacobi and Gauss-Seidel methods is diagonal dominance,
abs(a[i][i]) > sum( abs(a[i][j]), j=0...n-1, j!=i)
This test is also easy to implement as a check to run before the iteration.
The larger the relative gap in all these inequalities, the faster the convergence of the method.