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rabiulcste / Basic Template and Headers.cpp
Created January 7, 2016 15:37
Competitive Programming - Problem Solving Basic Codings
#include <bits//stdc++.h>
using namespace std;
#define MP make_pair
#define PB push_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define MOD 1000000007
void multiply(char a[], char b[], char res[])
{
int i, j, k, mul, carry;
// if one number is zero, then result obviously zero
if(strcmp(a, "0") == 0 || strcmp(b, "0") == 0)
{
strcpy(res, "0");
return;
}
void primeFactors(int n)
{
// Print the number of 2s that divide n
while (n%2 == 0)
{
printf("%d ", 2);
n = n/2;
}
// n must be odd at this point. So we can skip one element (Note i = i +2)
int main()
{
int n, cnt = 0;
cin >> n;
while(n)
{
n &= (n-1);
cnt++;
}
cout << cnt <<endl;
int countSetBits(unsigned int n)
{
unsigned int count = 0;
while(n)
{
count += n & 1; // ভাগশেষ ০ নাকি ১ । যদি ১ হয় তাহলে কাউন্টের মান বেড়ে যাবে ।
n >>= 1; // n কে দুই দিয়ে ভাগ করা হচ্ছে ।
}
return count;
}
ll totient (ll i)
{
ll res; /* Result */
ll j;
if (i==1) return 1;
res=i;
// Check for divisibility by every prime number below the square root.
bool isPrime(int n) {
if(n < 2) return false;
for(int i = 2; i < sqrt(n); i++)
if(n%i == 0) return false;
}
#define MAX 1000001
char prime[MAX]; // 0 দিয়ে initialize করতে হবে
void seive( int n ) // n পর্যন্ত প্রাইম বের করব
{
int x = sqrt( n );
prime[0] = prime[1] = 1; // 0 এবং 1 প্রাইম না
for( int i = 4; i <= n; i += 2 ) // জোড় সংখ্যাগুলোকে বাদ দিয়ে দিব
prime[i] = 1;
vector<char> vc;
char s1[1005] , s2[1005];
int dp[105][105];
void backtrace(int i, int j)
{
if(i == 0 || j == 0)
return;
else if(s1[i-1] == s2[j-1]) {
backtrace(i-1, j-1);
int main()
{
int n;
scanf("%d", &n);
map<string, int>mp;
for(int i=0; i<n; i++){
string s;
cin>>s;
if(mp[s] == 0) {mp[s]++; printf("OK\n");}
else {