Codeforces 474D Flowers (线性dp 找规律)

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).

题目连接：http://codeforces.com/problemset/problem/474/D

题目大意：一个东西爱吃花，有两种颜色红R和白W。他吃白花每次都一组一组吃，一组是连续在一起的k个，问在花的个数从ai到bi范围里。他总共同拥有多少种吃法

题目分析：dp[i]表示长度为i是他吃花的方案数。初始时dp[i] = 1 (0 <= i < k) i小于k时显然仅仅有一种

当i>=k时，我们dp[i]能够是dp[i - 1]加上一朵红花，或者dp[i - k]加上k朵白花，dp[i] = dp[i - 1] + dp[i - k]

然后求出dp数组的前缀和，查询时O(1)

#include <cstdio>
#include <cstring>
#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\n", (MOD + sum[b] - sum[a - 1]) % MOD);
    }
}