Flowers（DP）

D. Flowers

time limit per test

1.5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 modulo1000000007 (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).

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).

 

      题意：

      给出 t 和 k，代表有 t 组询问。k 代表一次只能吃 k 朵 W 花，后给出 t 组的 a - b 区间。问吃这段区间内的长度序列满足条件的方法数。

 

      思路：

      DP。dp [ i ] = dp [ i - k ] + dp [ i - 1] 代表一次要么连续吃 k 朵 W 花，要么一次吃 1 朵 R 花。记得最后答案中的减法后要 + MOD 再 % MOD。不然会出错。

 

      AC：

#include 
#include 
#include 

using namespace std;

typedef long long ll;

const ll MOD = 1000000007;
const int MAX = 100005;

ll dp[MAX];
ll sum[MAX];
ll a[MAX], b[MAX];

int main() {

    int t, k;
    scanf("%d%d", &t, &k);

    ll Max = 0;
    for (int i = 1; i <= t; ++i) {
        scanf("%I64d%I64d", &a[i], &b[i]);
        Max = max(Max, b[i]);
    }

    for (int i = 0; i < k; ++i) {
        dp[i] = 1;
        sum[i] = sum[i - 1] + dp[i];
    }

    for (int i = k; i <= Max; ++i) {
        dp[i] = (dp[i - k] + dp[i - 1]) % MOD;
        sum[i] = (dp[i] % MOD + sum[i - 1] % MOD) % MOD;
    }

    for (int i = 1; i <= t; ++i) {
        printf("%I64d\n", (sum[b[i]] - sum[a[i] - 1] + MOD) % MOD);
    }

    return 0;
}