Problems upto 35

18.py

@@ -0,0 +1,29 @@
+input_str = '''75
+95 64
+17 47 82
+18 35 87 10
+20 04 82 47 65
+19 01 23 75 03 34
+88 02 77 73 07 63 67
+99 65 04 28 06 16 70 92
+41 41 26 56 83 40 80 70 33
+41 48 72 33 47 32 37 16 94 29
+53 71 44 65 25 43 91 52 97 51 14
+70 11 33 28 77 73 17 78 39 68 17 57
+91 71 52 38 17 14 91 43 58 50 27 29 48
+63 66 04 68 89 53 67 30 73 16 69 87 40 31
+04 62 98 27 23 09 70 98 73 93 38 53 60 04 23'''
+str_list = input_str.split("\n")
+lst = [[int(j) for j in str_list[i].split()] for i in range(len(str_list))]
+ans_lst = lst
+str = len(lst)-2
+for r in range(str, -1, -1):
+  for c in range(0, len(lst[r])):
+    ans_lst[r][c] += max(lst[r+1][c], lst[r+1][c+1])
+print ans_lst[0][0]
+
+
+
+
+
+

21.py

@@ -0,0 +1,7 @@
+# sum of proper divisors
+def pds(n):
+    return sum(x for x in range(1, n//2+1) if not n%x)
+
+#print d(284)
+print sum(i for i in range(1, 10000) if pds(pds(i)) == i and i != pds(i))
+

23.py

@@ -0,0 +1,16 @@
+maxi = 28123+1
+# sum of proper divisors
+def pds(n):
+    return sum(x for x in range(1, n//2+1) if not n%x)
+
+O = set(range(1, maxi))
+#A set of abundant numbers, numbers are abundant if pds(n) > n
+A = set([n for n in range(1, maxi) if pds(n) > n])
+
+#B set of numbers which can be expressed as sum of two abundant numbers
+B = set([a1+a2 for a1 in A for a2 in A if a1+a2<=maxi])
+
+#C = O -B set of numbers which can't be expressed as sum of two abun nos. 
+C = O - B
+
+print sum(C)

24.py

@@ -0,0 +1,4 @@
+# print 1000000th permutation
+import itertools
+a = list[itertools.permutations('0123456789')]
+print a[999999]

25.py

@@ -0,0 +1,10 @@
+maxi = int('1%s' % ('0' * 999))
+st1 = 0
+st2 = 1
+cnt = 1
+while(len(str(st2))<=999):
+    curr, st1 = st1+st2, st2
+    st2 = curr
+    cnt += 1
+
+print cnt, len(str(st2))

26.py

@@ -0,0 +1,11 @@
+def get_decimals(num, den, stack =[]):
+    if num == 0 or den == 0:
+        return len(stack)
+    elif num in stack:
+        return len(stack) -1
+    else:
+        a,b = divmod(num, den)
+        return get_decimals(b*10, den, stack+[num])
+
+print  max((get_decimals(1, num), num) for num in range(2, 1001))
+

27.py

@@ -0,0 +1,50 @@
+import math
+
+def isPrime(n):
+	if(n%2 == 0):
+		return 0
+	for i in range(3,int(math.sqrt(n))+1):
+		if(n%i == 0):
+			return 0
+        return 1
+'''
+for n = 0, we should get a prime, we can ensure the largest prime, 
+by generating picking the max prime < 1000 hence b = 997
+primes = [i for i in range(2, 1000) if isPrime(i)]
+print primes[-1]
+hence the max we can go is 997
+
+now we simply put the equation as is and let my computer figure out the value. 
+'''
+
+primes = [i for i in range(2, 1000) if isPrime(i)]
+
+maxi = 997+1
+a_max = 0
+b_max = 0
+i_max = 0
+
+def get_prods(a, b):
+    global a_max
+    global b_max
+    global i_max
+    i = 0
+    while((i**2 + a*i + b) in primes):
+        i += 1
+        print a, b, i
+    if(i>i_max):
+        a_max = a
+        b_max = b
+        i_max = i
+    return 0
+
+for a in range(0, maxi):
+    for b in range(0, maxi):
+        get_prods(a, b)
+        get_prods(a, -b)
+        get_prods(-a, b)
+
+print a_max*b_max                
+
+
+

28.py

@@ -0,0 +1,43 @@
+'''Creating one more cycle of numbers, will show us an order,
+which will eventually reduce the computation which
+we need to perform. 
+
+the series from bottom left to top right:
+43 21 7 1 3 13 31
+re-arranged it looks like this: 1 3 7 13 21 31 43
+the difference between the difference of numbers increases by 2
+1+2
+3+4
+7+6
+13+8
+etc. 
+
+
+the series from top left to bottom right
+37 17 5 1 9 25 49
+re-arranged it looks like: 1 5 9 17 25 37 49
+the difference between the difference of numbers increases by 4 at a cycle length of 2
+1+4
+5+4
+
+9+8
+17+8
+
+25+12
+37+12
+etc. 
+'''
+s1 = s2 = 0;maxi = 1001;diff1 = 2;diff2 = 4;l1 = [1];l2 = [1]
+
+for cnt in range(1, maxi):
+    l2.append(l2[-1]+diff2)
+    if not (cnt&1): diff2+=4
+#print l2
+
+for cnt in range(1, maxi):
+    l1.append(l1[-1]+diff1)
+    diff1 += 2
+#print l1
+# removing an extra 1 since in both list1 and list2 one has occured. 
+print sum(l1) + sum(l2) - 1
+

29.py

@@ -0,0 +1,7 @@
+lst = [a**b for a in range(2, 101) for b in range(2, 101)]
+present = []
+for i in lst:
+    if not i in present:
+    present.append(i)
+print len(present)
+

30.py

@@ -0,0 +1,14 @@
+''' We need to figure out an upper bound, since it is not given in the question. Let the number of digits in such a number be 'd'
+(9^5)*d [A] will be the maximum sum of such a number.
+This maximum sum will always be less than 10^(d-1) [B]
+
+since A > B.
+we can find the above relation holds true for d<= 6
+hence we should go upto (9^5)*6
+
+'''
+
+high = 354294
+chksum = lambda num: sum(int(i)**5 for i in str(num))
+ans = sum(i for i in range(10, high+1) if i == chksum(i))
+print ans

31.py

@@ -0,0 +1,20 @@
+''' 
+This is the most famous coin change DP problem
+'''
+coins = [1, 2, 5, 10, 20, 50, 100, 200]
+maxi = 200
+lst = [1] + [0] * (maxi + 1)
+[lst.__setitem__(i, lst[i] + lst[i - c]) for c in coins for i in range(c, maxi+1)]
+print lst[200]
+
+
+'''
+BULLSHIT APPROACH
+for c in coins:
+    for i in range(c, maxi+1):
+        lst[i] += lst[i-c]
+'''
+
+
+
+

32.py

@@ -0,0 +1,35 @@
+'''
+The case is as follows:
+Master String = ABCDEFGHI
+x*y = z
+all the integers of x, y and z should contribute to 
+the master string 'uniquely'
+len(z) >= len(x) and len(z) >= len(y)
+Hence we come to the following conclusion that the solution can always be of the 
+form 
+1) ABC*DE = FGHI
+2) ABCD*E = FGHI
+NOT  AB*CD = EFGHI
+since 99 * 99 = 9801
+'''
+
+import itertools
+strn = "123456789"
+soln = []
+perms = list(itertools.permutations(strn))
+
+for perm in perms:
+    num = ''.join(list(perm))
+# processing form 1    
+    n1 = int(num[0:3]); n2 = int(num[3:5]); n3 = int(num[5:]);
+    if n1*n2 == n3 and n3 not in soln:
+        soln.append(n3)
+        print n1, n2, n3
+# processing second form (not the html form)
+    n1 = int(num[0:4]); n2 = int(num[4:5]); n3 = int(num[5:]);
+    if n1*n2 == n3 and n3 not in soln:
+        soln.append(n3)
+        print n1, n2, n3
+
+print sum(soln)
+

33.py

@@ -0,0 +1,19 @@
+num = 1;den = 1;
+for n in range(10, 100):
+    for d in range(n+1, 100):
+        if n%10 != 0 and d%10 != 0:
+            n = float(n); d = float(d)
+            num1, num2 = divmod(n, 10)
+            den1, den2 = divmod(d, 10)
+            if num2 == den1:
+                chk1 = num1/den2
+                chk2 = n/d
+                if chk1 == chk2:
+                    print n, d 
+                    num *= n; den *= d;
+
+
+print num, den
+
+
+

34.py

@@ -0,0 +1,9 @@
+maxi = 10000000
+facti = [1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880]
+ans = 0
+for i in range(10, maxi):
+    b = map(int, str(i))
+    if i == sum([facti[j] for j in b]):
+        ans += i
+        print i
+print ans        

35.py

@@ -0,0 +1,30 @@
+import math
+maxi = 1000000+1
+lst = range(2, maxi)
+ans = []
+
+
+def isPrime(n):
+	if(n%2 == 0):
+		return 0
+	for i in range(3,int(math.sqrt(n))+1):
+		if(n%i == 0):
+			return 0
+        return 1
+
+primes = [i for i in lst if isPrime(i)]
+print "primes generated"
+
+def chk_circ(num):
+    s = str(num)
+    perms = [int(s[i:] + s[:i]) for i in range(len(s))]
+    for numstr in perms:
+        if numstr not in primes:
+            return False
+    for numstr in perms:
+        ans.append(numstr)
+    return True
+[chk_circ(i) for i in primes if i not in ans]
+print ans  
+print len(ans)
+