Mr X is traveling by car on an expressway. Suppose there are several gas (petrol) stations
on the way: at distances 0 = d0 < d1 < d2 < ... < dn from the starting point d0.
Mr X’scar, when full, can travel a distance D >= max{di+1 - di} .
Mr X wants to minimize the number of stops he makes to fill gas.
Devise an greedy algorithm that return min numbers of stops needed.
This is the sort of thing it is asking for.
GCC 4.8.3: g++ -Wall -Wextra -std=c++0x -g main.cpp
#include <algorithm>
#include <iostream>
#include <stdexcept>
#include <vector>
int count_stops(int D, const std::vector<int>& d) {
if (d.size() < 2) { throw std::logic_error("Vector too small."); }
auto p = std::begin(d);
auto count = 0;
while (p != --std::end(d)) {
// Greedy: go as far as I can on this tank of gas.
auto n = --std::find_if(p, std::end(d), [=](int x) {
return *p + D < x; });
// The specification says we do not need to worry about this...
if (p == n) { throw std::logic_error("D too small."); }
p = n;
++count; }
return count; }
int main(int, char* []) {
auto D = 16;
auto d = std::vector<int> { 0, 5, 15, 30, 32, 33, 37, 49, 53, 59, 61 };
std::cout << "stops: " << count_stops(D, d) << "\n";
return 0; }
Related
I need to make a c program that will make a histogram of all the letters present in a phrase the user gives. When I run it, I does it but gives a "* stack smashing detected *: terminated". Where would this error be coming from? (for ease right now I set max to 3). In the future i'll have it find the max
Thank you
Andrew
#include <stdio.h>
#include <ctype.h>
#include <string.h>
static void ReadText(int histo[26],int max) {
char phrase[100];
int i;
char Letter;
char toArray;
// read in phrase
printf("Enter Phrase: "); // reads in phrase with spaces between words
scanf("%[^\n]",phrase);
// count the number of certain letters that occur
for(i = 0; i <= strlen(phrase);++i) {
Letter = phrase[i];
if(isalpha(Letter) != 0){
Letter = tolower(Letter);
toArray = Letter - 97;
histo[(int)toArray] = histo[(int)toArray] + 1;
}
}
}
static void DrawHist(int histo[26], int max){
int i;
int j;
int histo2[50];
for(i = 0; i <= 26; i++) {
histo2[i+i] = histo[i];
if(i < 25) {
histo2[i+i+1] = 0;
}
}
// (i = 1; i <= 50; i++) {
// printf("%d",histo2[i]);
//}
//printf("\n");
for(i=max;i>0;--i) {
for(j=0;j<=51;++j) {
if((j < 51) && (histo2[j] >= i)) {
printf("|");
}
else if((j < 51) && (histo2[j] < i)){
printf(" ");
}
else if(j == 51){
printf("\n");
}
}
}
printf("+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-\n");
printf("A B C D E F G H I J K L M N O P Q R S T U V W X Y Z\n");
}
int main() {
int histo[26] = {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};
int max = 3;
//int i;
ReadText(histo,max);
//for(i = 0; i<26;++i) {
// printf("%d",histo[i]);
//}
DrawHist(histo,max);
return 0;
}
Could anyone please help me understand why Clang reallocates the same memory address for different variables while their lifetimes intersect?
I am using a sample program (below) to show the problem.
When I compile the program with clang -O0, variable j in function ok has the same memory address as variable solutions in function nqueens.
Function ok is called inside function nqueens, which means that the lifetime of the variables intersect; the same stack space cannot be used/reused for both functions.
Compiling the program with gcc or clang at -O1, however, they are assigned different memory addresses.
Any help is appreciated!
#include <stdlib.h>
#include <stdio.h>
#include <memory.h>
#include <alloca.h>
/* Checking information */
static int solutions[] = {
1,
0,
0,
2,
10, /* 5 */
4,
40,
92,
352,
724, /* 10 */
2680,
14200,
73712,
365596,
};
#define MAX_SOLUTIONS sizeof(solutions)/sizeof(int)
int total_count;
int sharedVar = 0;
int ok(int n, char *a)
{
int i, j;
char p, q;
printf("jjjjjjjjj: %d, %p\n", n,&j);
for (i = 0; i < n; i++) {
p = a[i];
for (j = i + 1; j < n; j++) {
q = a[j];
if (q == p || q == p - (j - i) || q == p + (j - i))
return 0;
}
}
return 1;
}
void nqueens (int n, int j, char *a, int *solutions)
{
int i,res;
sharedVar = sharedVar * j - n;
if (n == j) {
/* good solution, count it */
*solutions = 1;
return;
}
printf("solutions: %d, %p\n", j, &solutions);
*solutions = 0;
/* try each possible position for queen <j> */
for (i = 0; i < n; i++) {
a[j] = (char) i;
if (ok(j + 1, a)) {
nqueens(n, j + 1, a,&res);
*solutions += res;
}
}
}
int main()
{
int size = 3;
char *a;
// printf("total_count: %p\n", &total_count);
total_count=0;
a = (char *)alloca(size * sizeof(char));
printf("Computing N-Queens algorithm (n=%d) ", size);
sharedVar = -5;
nqueens(size, 0, a, &total_count);
printf("completed!\n");
printf("sharedVar: %d\n", sharedVar);
}
I am trying to use Eigen unsupported FFT library using FFTW backend. Specifically I am want to do a 2D FFT. Here's my code :
void fft2(Eigen::MatrixXf * matIn,Eigen::MatrixXcf * matOut)
{
const int nRows = matIn->rows();
const int nCols = matIn->cols();
Eigen::FFT< float > fft;
for (int k = 0; k < nRows; ++k) {
Eigen::VectorXcf tmpOut(nRows);
fft.fwd(tmpOut, matIn->row(k));
matOut->row(k) = tmpOut;
}
for (int k = 0; k < nCols; ++k) {
Eigen::VectorXcf tmpOut(nCols);
fft.fwd(tmpOut, matOut->col(k));
matOut->col(k) = tmpOut;
}
}
I have 2 problems :
First, I get a segmentation fault when using this code on some matrix. This error doesn't happen for all matrixes. I guess it's related to an alignment error. I use the functions in the following way :
Eigen::MatrixXcf matFFT(mat.rows(),mat.cols());
fft2(&matFloat,&matFFT);
where mat can be any matrix. Funnily, the code plants only when I compute the FFT over the 2nd dimension, never on the first one. This doesn't happen with kissFFT backend.
Second I don't get the same result as Matlab (that uses FFTW), when the function works. Eg :
Input Matrix :
[2, 1, 2]
[3, 2, 1]
[1, 2, 3]
Eigen gives :
[ (0,5), (0.5,0.86603), (0,0.5)]
[ (-4.3301,-2.5), (-1,-1.7321), (0.31699,-1.549)]
[ (-1.5,-0.86603), (2,3.4641), (2,3.4641)]
Matlab gives :
17 + 0i 0.5 + 0.86603i 0.5 - 0.86603i
-1 + 0i -1 - 1.7321i 2 - 3.4641i
-1 + 0i 2 + 3.4641i -1 + 1.7321i
Only the central part is the same.
Any help would be welcome.
I failed to activate EIGEN_FFTW_DEFAULT in my first solution, activating it reveals an error in the fftw-support implementation of Eigen. The following works:
#define EIGEN_FFTW_DEFAULT
#include <iostream>
#include <unsupported/Eigen/FFT>
int main(int argc, char *argv[])
{
Eigen::MatrixXf A(3,3);
A << 2,1,2, 3,2,1, 1,2,3;
const int nRows = A.rows();
const int nCols = A.cols();
std::cout << A << "\n\n";
Eigen::MatrixXcf B(3,3);
Eigen::FFT< float > fft;
for (int k = 0; k < nRows; ++k) {
Eigen::VectorXcf tmpOut(nRows);
fft.fwd(tmpOut, A.row(k));
B.row(k) = tmpOut;
}
std::cout << B << "\n\n";
Eigen::FFT< float > fft2; // Workaround: Using the same FFT object for a real and a complex FFT seems not to work with FFTW
for (int k = 0; k < nCols; ++k) {
Eigen::VectorXcf tmpOut(nCols);
fft2.fwd(tmpOut, B.col(k));
B.col(k) = tmpOut;
}
std::cout << B << '\n';
}
I get this output:
2 1 2
3 2 1
1 2 3
(17,0) (0.5,0.866025) (0.5,-0.866025)
(-1,0) (-1,-1.73205) (2,-3.4641)
(-1,0) (2,3.4641) (-1,1.73205)
Which is the same as your Matlab result.
N.B.: FFTW seems to support 2D real->complex FFT natively (without using individual FFTs). This is likely more efficient.
fftwf_plan fftwf_plan_dft_r2c_2d(int n0, int n1,
float *in, fftwf_complex *out, unsigned flags);
I want to reduce runtime of my code by use the OpenACC but unfortunately when i use OpenACC the output becomes zero.
sajad.**
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <assert.h>
#include <openacc.h>
#include<time.h>
#include <string.h>
#include <malloc.h>
#define NX 201
#define NY 101
#define NZ 201
int main(void)
{
int i, j, k, l, m;
static double tr, w;
static double dt = 9.5e-9, t;
static double cu[NZ];
static double AA[NX][NY][NZ] , CC[NX][NY][NZ] , BB[NX][NY][NZ] ;
static double A[NX][NY][NZ] , B[NX][NY][NZ] , C[NX][NY][NZ] ;
FILE *file;
file = fopen("BB-and-A.csv", "w");
t = 0.;
#pragma acc data copyin( tr, w,dt, t),copy(B ,A , C,AA , CC,BB,cu )
{
for (l = 1; l < 65; l++) {
#pragma acc kernels loop private(i, j,k)
for (i = 1; i < NX - 1; i++) {
for (j = 0; j < NY - 1; j++) {
for (k = 1; k < NZ - 1; k++) {
A[i][j][k] = A[i][j][k]
+ 1. * (B[i][j][k] - AA[i][j][k - 1]);
}
}
}
#pragma acc kernels loop private(i, j,k)
for (i = 1; i < NX - 1; i++) { /* BB */
for (j = 1; j < NY - 1; j++) {
for (k = 0; k < NZ - 1; k++) {
B[i][j][k] = B[i][j][k]
+ 1.* (BB[i][j][k] - A[i - 1][j][k]);
}
}
}
#pragma acc kernels
for (m = 1; m < NZ - 1; m++) {
tr = t - (double)(m)*5 / 1.5e8;
if (tr <= 0.)
cu[m] = 0.;
else {
w = (tr / 0.25e-6)*(tr / 0.25e-6);
cu[m] =1666*w / (w + 1.)*exp(-tr / 2.5e-6) ;
cu[m] = 2*cu[m];
}
A[10][60][m] = -cu[m];
}
#pragma acc update self(B)
fprintf(file, "%e, %e \n", t*1e6, -B[22][60][10] );
t = t + dt;
}
}
fclose(file);
}
The problem here is the "copyin( tr, w,dt, t)", and in particular the "t" variable. By putting these scalars in a data clause, you'll need to managed the synchronization between the host as device copies. Hence, when you update the variable on the host (i.e. "t = t + dt;"), you then need to update the device copy with the new value.
Also, there's a potential race condition on "tr" since the device code will now the shared device variable instead of a private copy.
Though, the easiest thing to do is to simply not put these scalars in a data clause. By default, OpenACC privatizes scalars so there's no need manage them yourself. In t's case, it's value will be passed as an argument to the CUDA kernel.
To fix your code change:
#pragma acc data copyin( tr, w,dt, t),copy(B ,A , C,AA , CC,BB,cu )
to:
#pragma acc data copy(B ,A , C,AA , CC,BB,cu )
Note that there's no need to put the loop indices in a private clause since they are implicitly private.
I want to know if A and B are relatively prime using Euclidean Algorithm. A and B are large numbers that cannot be stored in any data type(in C), so they are stored in a linked list. In the algorithm, the operator % is used. My question is, is there a way to compute for A mod B without actually directly using the % operator. I found out that % is distributive over addition:
A%B = ((a1%B)+(a2%B))%B.
But the problem still persists because I will still be doing %B operations.
You need calculate a % b without the % operator. OK? By definition the modulo operation finds the remainder after division of one number by another.
In python:
# mod = a % b
def mod(a, b):
return a-b*int(a/b)
>>> x = [mod(i,j) for j in range(1,100) for i in range(1,100)]
>>> y = [i % j for j in range(1,100) for i in range(1,100)]
>>> x == y
True
In C++:
#include <iostream>
#include <math.h>
using namespace std;
unsigned int mod(unsigned int a, unsigned int b) {
return (unsigned int)(a-b*floor(a/b));
}
int main() {
for (unsigned int i=1; i<=sizeof(unsigned int); ++i)
for (unsigned int j=1; j<=sizeof(unsigned int); ++j)
if (mod(i,j) != i%j)
cout << "Somthing wrong!!";
cout << "Proved for all unsigned int!";
return 0;
}
Proved for all unsigned int!
Now, just extend the result to your big numbers...!!!