Flowers (Codeforces 474D)
Tags: DP, dp

We saw the little game Marmot made for Mole's lunch. Now it's Marmot's dinner time and, as we all know, Marmot eats flowers. At every dinner he eats some red and white flowers. Therefore a dinner can be represented as a sequence of several flowers, some of them white and some of them red.But, for a dinner to be tasty, there is a rule: Marmot wants to eat white flowers only in groups of size k.Now Marmot wonders in how many ways he can eat between a and b flowers. As the number of ways could be very large, print it modulo 1000000007 (109 + 7).

Input: Input contains several test cases.The first line contains two integers t and k (1 ≤ t, k ≤ 105), where t represents the number of test cases.The next t lines contain two integers ai and bi (1 ≤ ai ≤ bi ≤ 105), describing the i-th test.

Output: Print t lines to the standard output. The i-th line should contain the number of ways in which Marmot can eat between ai and bi flowers at dinner modulo 1000000007 (109 + 7).

Note:   For K = 2 and length 1 Marmot can eat (R).  For K = 2 and length 2 Marmot can eat (RR) and (WW).  For K = 2 and length 3 Marmot can eat (RRR), (RWW) and (WWR).  For K = 2 and length 4 Marmot can eat, for example, (WWWW) or (RWWR), but for example he can't eat (WWWR). 

Sample Inputs:
3 2
1 3
2 3
4 4

Sample Outputs:
6
5
5

Solution (Python):
def R():
    return map(int, input().split())
(a, b) = R()
t = ''
n = 100001
k = 10 ** 9 + 7
A = list(range(1, b + 1))
for _ in range(b, n):
    A += [(A[-1] + A[-b] + 1) % k]
for _ in range(a):
    (m, n) = R()
    t += str((A[n] - A[m - 1]) % k) + '\n'
print(t)

Solution (Python):
def R():
    return map(int, input().split())
(a, b) = R()
n = 100001
s = ''
k = 10 ** 9 + 7
A = [1] * b
B = list(range(1, b + 1))
for i in range(b, n):
    t = A[-1] + A[-b]
    A += [t % k]
    B += [(B[-1] + t) % k]
for _ in range(a):
    (m, n) = R()
    s += str((B[n] - B[m - 1]) % k) + '\n'
print(s)

Solution (Python):
import sys
import sys, threading
from heapq import heapify, heappop, heappush
from math import log10, floor, pow, gcd, sqrt, inf, log2
from collections import defaultdict, deque, Counter
input = sys.stdin.readline

def inp():
    return int(input())

def inlt():
    return list(map(int, input().split()))

def insr():
    s = input()
    return list(s[:len(s) - 1])

def invr():
    return map(int, input().split())

def main():
    MOD = 1000000007
    N = 100005

    def solve(n, k, test_cases):
        dp = [0] * N
        i = 1
        while i < k:
            dp[i] = 1
            i += 1
        dp[k] = 2
        for i in range(k + 1, N):
            dp[i] = (dp[i - k] + dp[i - 1]) % MOD
        for i in range(1, N):
            dp[i] = (dp[i] + dp[i - 1]) % MOD
        result = []
        for (a, b) in test_cases:
            k = (dp[b] - dp[a - 1]) % MOD
            if k < 0:
                k += MOD
            result.append(k)
        return result
    (n, k) = map(int, input().split())
    test_cases = []
    for _ in range(n):
        (a, b) = map(int, input().split())
        test_cases.append((a, b))
    output = solve(n, k, test_cases)
    for ways in output:
        print(ways)
threading.stack_size(1 << 27)
sys.setrecursionlimit(1 << 30)
main_thread = threading.Thread(target=main)
main_thread.start()
main_thread.join()