krisys

#include<iostream>
using namespace std;
int sudoku[9][9],ptr[162],xstart[9]={0,3,6,0,3,6,0,3,6}, ystart[9]={0,0,0,3,3,3,6,6,6};
int valid(){
    int valid=1;
    int hashtable[10];
    for(int i=0;i<9;i++){
        for(int j=0;j<9;j++){
            for(int k=j+1;k<9;k++){
	        if(sudoku[i][j]==0) continue;

krisys

#include<iostream>
#include<string>
#include<vector>
#define SIZE 4
#define debug(x) cout<< #x << " = "<< x <<endl
#define swap(typ,A,B) { typ temp;temp=B; B=A;A=temp;}
using namespace std;

vector <int> permset;
string flags;

krisys

#include<iostream>
#include<string>
using namespace std;

string deleteKthCharacter(string str, int k){
    return str.erase(k, 1);
}

void permute( string dest, string source, int len){
    if( dest.length() == len){

krisys

#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;

int main(){
    int n;
    char c;
    vector < string > next, current;

krisys

#include<iostream>
#include<string>
#define debug(x) cout<< #x << " = "<< x <<endl
#define swap(typ,A,B) { typ temp;temp=B; B=A;A=temp;}
using namespace std;

class Node{
    friend class HuffmanTree;
    char ch;
    float prob;

krisys

# Problem 1
"""Find the sum of all the multiples of 3 or 5 below 1000."""

sum( [ num for num in range(3,1000) if num % 3 == 0 or num % 5 == 0 ] )


# Problem 4
"""
A palindromic number reads the same both ways. The largest palindrome 
made from the product of two 2-digit numbers is 9009 = 91  99.

krisys

====================================================================
Simple-as-possible instructions to add a field (or more) using South 
to an existing Django model with existing data.
====================================================================

Two versions:
  1. Super-condensed (the bare minimum - jfdi)
  2. Detailed-but-brief (if you want more information).

Notes:

krisys

char *token = strtok(char *str, char *delim); // first time.

while(token != NULL){
    token = strtok(NULL, delim); // to retrieve the subsequent tokens.
}

krisys

/* thread 1 */

char *str1 = "1,2,3", *token;
token = strtok(str1, ","); 

while(token != NULL){
    token = strtok(NULL, delim); 
}

/* thread 2 */

krisys

string RainyRoad::isReachable(vector <string> road) {
    int n = road[0].length();
    if(road[0][0] == 'W' || road[0][n-1] == 'W')
        return "NO";
    for(int i = 0 ;i < n; i++){
        if(road[0][i] == 'W' && road[1][i] == 'W')
            return "NO";
    }
    return "YES";	
}