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‎10001stprime.py‎

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'''
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Created on May 8, 2015
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@author: Chocolate
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'''
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def SieveofEras(arr):
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arr[0]=True;
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arr[1]=True;
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for i in xrange(1,len(arr)):
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if arr[i]==False:
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for j in xrange(i*2,len(arr),i):
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arr[j]=True;
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return arr;
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ll=[False]*10002
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ll=SieveofEras(ll);
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for _ in xrange(input()):
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n=input();
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idx=2;
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while idx<len(ll):
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#print idx;
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if(n==0):
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break;
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if(ll[idx]==False):
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n-=1;
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idx+=1;
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print(idx-1);

‎BSLArgestPalinDromProduct.py‎

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'''
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Created on May 7, 2015
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https://www.hackerrank.com/contests/projecteuler/challenges/euler004
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@author: Chocolate
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'''
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def binSearch(arr,elm):
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lo=0;
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hi=len(arr)-1;
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#print hi;
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while(lo<=hi):
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mid=lo+((hi-lo)/2);
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#print mid
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if(arr[mid]==elm):
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return mid;
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elif(arr[mid]>elm):
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hi=mid-1;
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else:
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lo=mid+1;
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#print lo
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return hi;
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ll=[];
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for i in range(100,1000):
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for j in range(100,1000):
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mulnum=i*j;
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mulnums=str(mulnum);
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if(mulnums==mulnums[::-1]):
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ll.append(mulnum);
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ll=sorted(ll);
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#print ll;
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for _ in xrange(input()):
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n=input();
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print(ll[binSearch(ll, n)]);
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‎CoinSum.py‎

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'''
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Created on May 6, 2015
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int target = 200;
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int[] coinSizes = { 1, 2, 5, 10, 20, 50, 100, 200 };
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int[] ways = new int[target+1];
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ways[0] = 1;
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for (int i = 0; i < coinSizes.Length; i++) {
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for (int j = coinSizes[i]; j <= target; j++) {
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ways[j] += ways[j - coinSizes[i]];
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}
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}
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https://www.hackerrank.com/contests/projecteuler/challenges/euler031
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http://www.mathblog.dk/project-euler-31-combinations-english-currency-denominations/
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@author: Chocolate
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'''
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for _ in xrange(input()):
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n=input();
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coinSizes=[ 1, 2, 5, 10, 20, 50, 100, 200 ];
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ways=[0]*(n+1);
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ways[0]=1;
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for i in coinSizes:
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j=i;
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while j<=n:
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ways[j]+=ways[j - i];
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j+=1;
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print(ways[n]%(10**9+7));
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‎Compo.py‎

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'''
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Created on May 10, 2015
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@authorr: Chocolate
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'''
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from math import sqrt;
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class MyClass(object):
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def __init__(self, real=0.0,imag=0.0):
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self.real=real;
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self.imag=imag;
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def __add__(self, x):
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r=self.real+x.real;
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i=self.imag+x.imag;
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return MyClass(r,i);
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def __sub__(self, x):
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r=self.real-x.real;
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i=self.imag-x.imag;
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return MyClass(r,i);
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def __mul__(self, x):
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r=(self.real*x.real)-(self.imag*x.imag);
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i=(self.real*x.imag)+(self.imag*x.real);
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return MyClass(r,i);
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def __div__(self, x):
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'''
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float div=(x.real*x.real) + (x.imag*x.imag);
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complex tmp;
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tmp.real=(real*c.real)+(imag*c.imag);
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tmp.real/=div;
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tmp.imag=(imag*c.real)-(real*c.imag);
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tmp.imag/=div;
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return tmp;
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'''
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d=(x.real*x.real) + (x.imag*x.imag);
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r=(self.real*x.real)+(self.imag*x.imag);
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r=r/d;
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i=(self.imag*x.real)-(self.real*x.imag);
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i=i/d;
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return MyClass(r,i);
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def modu(self):
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z=(self.real*self.real)+(self.imag*self.imag);
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return sqrt(z);
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def __str__(self):
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if(self.imag<0):
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#print "%.2f %.2f" % (self.real, self.imag),"i"
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return str("%.2f - %.2fi" % (self.real, abs(self.imag)))
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#print "%.2f + %.2f" % (self.real, self.imag),"i"
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return str("%.2f + %.2fi" % (self.real, self.imag))
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n1,n2=map(float,raw_input().split());
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n3,n4=map(float,raw_input().split());
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a=MyClass(n1+0.0,n2+0.0);
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b=MyClass(n3+0.0,n4+0.0);
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print(a+b);
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print(a-b);
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print(a*b);
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print(a/b);
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print(round(a.modu(),2));
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print(round(b.modu(),2));
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#print a+b;

‎Counting_Fractions.py‎

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'''
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Created on May 6, 2015
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https://www.hackerrank.com/contests/projecteuler/challenges/euler072
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@author: Chocolate
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'''
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def P(L):
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phi = range(L+1)
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for n in xrange(2, L+1):
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if phi[n] == n:
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for k in xrange(n, L+1, n):
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phi[k] -= phi[k] // n
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return sum(phi) - 1
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for _ in xrange(input()):
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n=input();
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print(P(n));

‎EvenFibonaccinumbers.py‎

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'''
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Created on May 6, 2015
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@author: Chocolate
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'''
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import abc
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'''
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When taking the following rules into account:
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even + even = even
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even + odd = odd
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odd + even = odd
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odd + odd = even
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The parity of the first Fibonacci numbers is:
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o o e o o e o o e ...
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Thus basically, you simply need to do steps of three. Which is:
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(1,1,2)
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(3,5,8)
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(13,21,34)
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Given (a,b,c) this is (b+c,b+2*c,2*b+3*c).
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This means we only need to store the two last numbers, and calculate given (a,b), (a+2*b,2*a+3*b).
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Thus (1,2) -> (5,8) -> (21,34) -> ... and always return the last one.
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This will work faster than a "filter"-approach because that uses the if-statement which reduces pipelining.
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def checkTheMultiple(arr, num, n):
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# print("In checker");
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i = 2;
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a = num * i;
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while a <= n:
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# print(a);
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arr[a - 1] = 1;
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# print(arr);
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i = i + 1;
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a = num * i;
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def checkPrime(n):
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resultlist=[];
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if n >= 2:
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llist = [0] * n;
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for j in xrange(1, n, 1):
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if llist[j] == 0:
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checkTheMultiple(llist, j + 1, n);
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# print("Back");
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# print(llist);
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resultlist.extend([j+1]);
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return resultlist;
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'''
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def fibo(n):
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a=[0]*(n+1);
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a[0]=1;
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a[1]=2;
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for i in xrange(2,n+1,2):
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a[i]=a[i-2]+(a[i-1]*2);
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a[i+1]=(2*a[i-2])+(a[i-1]*3);
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if a[i+1]>=n:
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break;
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return a;
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for _ in xrange(input()):
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N=input();
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abc=[x for x in fibo(N) if x % 2 == 0];
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#print fibo(N);
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bca=[x for x in abc if x<N];
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#print abc;
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print(sum(bca));
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‎LargestPrimeFactor.py‎

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'''
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Created on May 7, 2015
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https://www.hackerrank.com/contests/projecteuler/challenges/euler003
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@author: Chocolate
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'''
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def prime(n):
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i = 2
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#print n;
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while (n % i != 0 and i*i< n):
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i += 1
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#print "In While",i
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if (i*i < n):
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#print "In if",i;
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return prime (n / i)
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else:
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print n
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for _ in xrange(input()):
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n = input()
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prime(n)

‎Multiplesof3and5.py‎

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'''
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Created on May 6, 2015
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https://www.hackerrank.com/contests/projecteuler/challenges/euler001
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@author: Chocolate
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'''
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def sum1(n,d):
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m = int(n/d);
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return (d*m*(m+1)/2);
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for _ in xrange(input()):
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n=input();
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print(sum1(n-1,3)+sum1(n-1,5)-sum1(n-1,15));
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'''
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def check15(num):
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if num%15==0:
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return 1;
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else:
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return 0;
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for _ in xrange(input()):
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n=input();
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a=3;
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b=5;
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i=a;
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j=b;
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sum1=0;
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count=2;
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var=1;
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if n<3:
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print(0);
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continue;
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while var==1:
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if (i>=n) and (j>=n):
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break;
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if(i<n):
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sum1+=i;
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i=a*count;
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if(j<n):
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sum1+=j;
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j=b*count;
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count+=1;
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a=15;
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i=a;
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count=2;
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while var==1:
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if i>=n:
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break;
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sum1=sum1-i;
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i=a*count;
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count+=1;
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print(sum1);
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'''
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'''def checkDiv(num):
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if (num%3==0) or (num%5==0):
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return 1;
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else:
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return 0;
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for _ in xrange(input()):
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n=input();
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sum1=0;
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for i in xrange(3,n,1):
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if(checkDiv(i)==1):
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sum1+=i;
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print(sum1);
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'''
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‎RangeLCM.py‎

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'''
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Created on May 7, 2015
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@author: Chocolate
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'''
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def RangeLCM(first, last):
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factors = range(first, last+1)
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for i in range(0, len(factors)):
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if factors[i] != 1:
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n = first + i
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for j in range(2*n, last+1, n):
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factors[j-first] = factors[j-first] / factors[i]
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return reduce(lambda a,b: a*b, factors, 1)

‎RangeLCM.pyc‎

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