Codeforces 474D Flowers (線性dp 找規律)

Codeforces 474D Flowers (線性dp 找規律)

 

D. Flowers time limit per test：1.5 seconds memory limit per test：256 megabytes

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 a_i and b_i (1 ≤ a_i ≤ b_i ≤ 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 a_i and b_i flowers at dinner modulo 1000000007 (10⁹ + 7).

Sample test(s) Input

3 2
1 3
2 3
4 4

Output

6
5
5

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).

 

 

題目大意：一個東西愛吃花，有兩種顏色紅R和白W，他吃白花每次都一組一組吃，一組是連續在一起的k個，問在花的個數從ai到bi范圍裡，他總共有多少種吃法

題目分析：dp[i]表示長度為i是他吃花的方案數，初始時dp[i] = 1 (0 <= i < k) i小於k時顯然只有一種

當i>=k時，我們dp[i]可以是dp[i - 1]加上一朵紅花，或者dp[i - k]加上k朵白花，dp[i] = dp[i - 1] + dp[i - k]

然後求出dp數組的前綴和，查詢時O(1)

 

 

#include 
#include 
#define ll long long
int const MAX = 1e5 + 5;
int const MOD = 1e9 + 7;
int a[MAX];
ll dp[MAX], sum[MAX];

int main()
{
    int t, k;
    scanf(%d %d, &t, &k);
    for(int i = 0; i < k; i++)
        dp[i] = 1;
    for(int i = k; i < MAX; i++)
        dp[i] = (dp[i - 1] % MOD + dp[i - k] % MOD) % MOD;
    sum[1] = dp[1];
    for(int i = 2; i < MAX; i++)
        sum[i] = (sum[i - 1] % MOD + dp[i] % MOD) % MOD;
    while(t --)
    {
        int a, b;
        scanf(%d %d, &a, &b);
        printf(%lld
, (MOD + sum[b] - sum[a - 1]) % MOD);
    }
}