Problem:


Score : 300 points

Problem Statement

In a public bath, there is a shower which emits water for T seconds when the switch is pushed.

If the switch is pushed when the shower is already emitting water, from that moment it will be emitting water for T seconds.
Note that it does not mean that the shower emits water for T additional seconds.

N people will push the switch while passing by the shower.
The i-th person will push the switch t_i seconds after the first person pushes it.

How long will the shower emit water in total?

Constraints

1 ≤ N ≤ 200,000

1 ≤ T ≤ 10^9

0 = t_1 < t_2 < t_3 < , ..., < t_{N-1} < t_N ≤ 10^9

T and each t_i are integers.

Input

Input is given from Standard Input in the following format:

N T
t_1 t_2 ... t_N

Output

Assume that the shower will emit water for a total of X seconds. Print X.

Sample Input 1

2 4
0 3

Sample Output 1

7

Three seconds after the first person pushes the water, the switch is pushed again and the shower emits water for four more seconds, for a total of seven seconds.

Sample Input 2

2 4
0 5

Sample Output 2

8

One second after the shower stops emission of water triggered by the first person, the switch is pushed again.

Sample Input 3

4 1000000000
0 1000 1000000 1000000000

Sample Output 3

2000000000

Sample Input 4

1 1
0

Sample Output 4

1

Sample Input 5

9 10
0 3 5 7 100 110 200 300 311

Sample Output 5

67



Sample input:
2 4
0 3


Sample output:
7

Submission status: Runtime Error

Solution code:
import collections
import functools
import time

class memoized(object):
    def __init__(self, func):
       self.func = func
       self.cache = {}

    def __call__(self, *args):
       if not isinstance(args, collections.Hashable):
          return self.func(*args)

       if args in self.cache:
          return self.cache[args]

       else:
          value = self.func(*args)
          self.cache[args] = value
          return value

    def __repr__(self):
       return self.func.__doc__

    def __get__(self, obj, objtype):
       return functools.partial(self.__call__, obj)

def knapsack(items, maxweight):
    @memoized
    def bestvalue(i, j):
        if i == 0: return 0
        value, weight = items[i - 1]
        if weight > j:
            return bestvalue(i - 1, j)
        else:
            return max(bestvalue(i - 1, j),
                       bestvalue(i - 1, j - weight) + value)

    j = maxweight
    result = []
    for i in xrange(len(items), 0, -1):
        if bestvalue(i, j) != bestvalue(i - 1, j):
            result.append(items[i - 1])
            j -= items[i - 1][1]
    result.reverse()
    return bestvalue(len(items), maxweight), result

def input_str(row):
    res = []
    for _ in range(row):
        res.append(raw_input().split())
    return res

N, W = [int(i) for i in raw_input().split()]
wvlist = input_str(N)
start = time.time()
items = [(int(v), int(w)) for w, v in wvlist]
print knapsack(items, W)[0]
print time.time() - start

Submission status: Runtime Error

Solution code:
import collections
import functools

class memoized(object):
    '''Decorator. Caches a function's return value each time it is called.
    If called later with the same arguments, the cached value is returned
    (not reevaluated).
    '''
    def __init__(self, func):
       self.func = func
       self.cache = {}
    def __call__(self, *args):
       if not isinstance(args, collections.Hashable):
          # uncacheable. a list, for instance.
          # better to not cache than blow up.
          return self.func(*args)
       if args in self.cache:
          return self.cache[args]
       else:
          value = self.func(*args)
          self.cache[args] = value
          return value
    def __repr__(self):
       '''Return the function's docstring.'''
       return self.func.__doc__
    def __get__(self, obj, objtype):
       '''Support instance methods.'''
       return functools.partial(self.__call__, obj)

def knapsack(items, maxweight):
    """
    Solve the knapsack problem by finding the most valuable
    subsequence of `items` subject that weighs no more than
    `maxweight`.

    `items` is a sequence of pairs `(value, weight)`, where `value` is
    a number and `weight` is a non-negative integer.

    `maxweight` is a non-negative integer.

    Return a pair whose first element is the sum of values in the most
    valuable subsequence, and whose second element is the subsequence.

    >>> items = [(4, 12), (2, 1), (6, 4), (1, 1), (2, 2)]
    >>> knapsack(items, 15)
    (11, [(2, 1), (6, 4), (1, 1), (2, 2)])
    """

    # Return the value of the most valuable subsequence of the first i
    # elements in items whose weights sum to no more than j.
    @memoized
    def bestvalue(i, j):
        if i == 0: return 0
        value, weight = items[i - 1]
        if weight > j:
            return bestvalue(i - 1, j)
        else:
            return max(bestvalue(i - 1, j),
                       bestvalue(i - 1, j - weight) + value)

    j = maxweight
    result = []
    for i in xrange(len(items), 0, -1):
        if bestvalue(i, j) != bestvalue(i - 1, j):
            result.append(items[i - 1])
            j -= items[i - 1][1]
    result.reverse()
    return bestvalue(len(items), maxweight), result

def input_str(row):
    res = []
    for _ in range(row):
        res.append(raw_input().split())
    return res

N, W = [int(i) for i in raw_input().split()]
wvlist = input_str(N)
items = [(int(v), int(w)) for w, v in wvlist]
print knapsack(items, W)[0]