codeforces 474D Flowers dp

传送门:cf 474d

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 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).

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).

        吃(?)红花与白花,吃白花有个条件,必须k朵白花一起吃,问这种情况下吃a朵花到b朵花之间一共有多少吃(!!)法。

        枚举出k的情况下,前10朵花吃的方法数就能发现,在k的条件下,a[i] = a[i-1] + a[i-k] (a[0]==1,i-k<0 a[i-k]=0) a[i]表示吃i朵花的方法数

        依据该转移方程求出前l朵花的方法数总和,就能快速求得a～b的方法数总和

/******************************************************
 * File Name:   d.cpp
 * Author:      kojimai
 * Creater Time:2014年10月07日 星期二 00时01分01秒
******************************************************/

#include
#include
#include
#include
#include
using namespace std;
#define mod 1000000007
#define FFF 100005
int a[FFF],sum[FFF];
struct node
{
	int a,b;
}p[FFF];
void init(int r,int k)
{
	a[0] = 1;
	for(int i = 1;i < k&&i <= r;i++)
	{
		a[i] = 1;
		sum[i] = (sum[i-1]+a[i])%mod;
	}
	for(int i = k;i <= r;i++)
	{
		a[i] = (a[i-1]+a[i-k])%mod;
		sum[i] = (sum[i-1]+a[i])%mod;
	}
}
int main()
{
	int t,k;
	scanf("%d%d",&t,&k);
	sum[0]=0;
	int Max=0;
	for(int i=0;i