How to write programme for 'M' for loops dynamically

Replies

• 

  sookie

  Hi spp,
  
  Sorry but your problem is not clear to me. Below is my trial of what I understood from the problem specified in above post.
  
  

  package myjava;
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  public class MTimesForLoop {
      public static void main(String[] args) throws IOException{
                  BufferedReader bin = new BufferedReader(new InputStreamReader(System.in));
                  int M= Integer.parseInt(bin.readLine());
                  for(int i=0;iOutput:
  
  3
  
  0 loop out of 3 specified loops
  
  1 loop out of 3 specified loops
  
  2 loop out of 3 specified loops
  Please add exactly what kind of stuff you are looking for? Also, feel free to correct if anything wrong done.
  
   
  
  Thanks !
                                      

• 

  spp

  sookie

  Hi spp,
  
  Sorry but your problem is not clear to me. Below is my trial of what I understood from the problem specified in above post.
  
  

  package myjava;
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  public class MTimesForLoop {
      public static void main(String[] args) throws IOException{
                  BufferedReader bin = new BufferedReader(new InputStreamReader(System.in));
                  int M= Integer.parseInt(bin.readLine());
                  for(int i=0;iOutput:
  
   
  
   
  
  Please add exactly what kind of stuff you are looking for? Also, feel free to correct if anything wrong done.
  
   
  
  Thanks !

  HI sookie
  I think u dont understand my problem.what problem I am facing is for any value of 'M' total 'M' for loops will execute.
  example:
  let i want to print "Hello world" and M=3
  then the code will be
  for(i=0;i {
  for(j=0;j {
  for(k=0;k {
  printf("Hello world");
  }
  }
  }
  for M=4
  total 4 for loops will be used ie
  for(i=0;i {
  for(j=0;j {
  for(k=0;k {
  for(l=0;l {
  printf("Hello world");
  }
  }
  }
  }
• 

  pradeep_agrawal

  Below is my code for the given problem statement solved using recursion.
  
  

  #include "stdio.h"
  
  void function(int m, int n, int* loopIndexes) {
    printf("Hello\n");
  }
  
  
  void mNumberOfLoops(int m, int n, int currentLoop, int* loopIndexes) {
    int i = 0;
  
    if(currentLoop != m) {
      for(i = 0; i < n; i++) {
        loopIndexes[currentLoop] = i;
        mNumberOfLoops(m, n, currentLoop + 1, loopIndexes);
      }
    } else {
      function(m, n, loopIndexes);
    }
  }
  
  
  int main() {
    int m = 0, n = 0;
    int *loopIndexes = NULL;
    int i = 0;
  
    scanf("%d%d", &m, &n);
    if((m < 1) || (n < 1)) {
      printf("Invalid input\n");
      return 0;
    }
  
    loopIndexes = (int*)malloc(sizeof(int)*m);
    for(i = 0; i < m; i++) {
      loopIndexes[i] = 0;
    }
  
    mNumberOfLoops(m, n, 0, loopIndexes);
  
    return 0;
  }
  

  Let me know if you have some query.
  
  -Pradeep
• 

  sookie

  @spp Oh so you were talking about "nested for loops -m times". Dude if you simply want to print("Hello") M[sup]M[/sup] times then why so much complications with for loops? Are you planning to do something else also in each of your "nested for loop" everytime ?
• 

  spp

  sookie

  @spp Oh so you were talking about "nested for loops -m times". Dude if you simply want to print("Hello") M[sup]M[/sup] times then why so much complications with for loops? Are you planning to do something else also in each of your "nested for loop" everytime ?

  yah dude.m writing a programme to find the basic feasible solutionof a given system of equation.
  for that i have to select a m*m matrix from a given m*n matrix and for that i need the use of 'nested for loops'.
  thanks dude for showing interest on this problem.
• 

  spp

  pradeep_agrawal

  Below is my code for the given problem statement solved using recursion.
  
  

  #include "stdio.h"
  
  void function(int m, int n, int* loopIndexes) {
    printf("Hello\n");
  }
  
  
  void mNumberOfLoops(int m, int n, int currentLoop, int* loopIndexes) {
    int i = 0;
  
    if(currentLoop != m) {
      for(i = 0; i < n; i++) {
        loopIndexes[currentLoop] = i;
        mNumberOfLoops(m, n, currentLoop + 1, loopIndexes);
      }
    } else {
      function(m, n, loopIndexes);
    }
  }
  
  
  int main() {
    int m = 0, n = 0;
    int *loopIndexes = NULL;
    int i = 0;
  
    scanf("%d%d", &m, &n);
    if((m < 1) || (n < 1)) {
      printf("Invalid input\n");
      return 0;
    }
  
    loopIndexes = (int*)malloc(sizeof(int)*m);
    for(i = 0; i < m; i++) {
      loopIndexes[i] = 0;
    }
  
    mNumberOfLoops(m, n, 0, loopIndexes);
  
    return 0;
  }
  

  Let me know if you have some query.
  
  -Pradeep

  thanks man...