CF#271 (Div. 2) D 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：在一个区间里面吃红白两种颜色的花，当长度为n时，要么连续吃k个白花，要么吃n个红花

状态分析：

对于每一次吃，有两种选择，要么吃k个白花，则这一次总共有dp[i-k]种吃法，要么吃一个红花，那么总共有dp[i-1]种吃法，则状态转移方程就是dp[i]=dp[i-k](这一次吃k个白）+dp[i-1]（这一次吃一个红） 

#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#define mod 1000000007
using namespace std;

__int64 dp[100009];
__int64 sum[100009];


int main()
{
    int k,t;
   
    cin>>t>>k;
        memset(dp,0,sizeof dp);
        memset(sum,0,sizeof sum);

        for(int i=0;i<=100001;i++)
        {
            if(i<k)
            {
                dp[i]=1;
                sum[i]=sum[i-1]+dp[i];
            }
            else
            {
                dp[i]=dp[i-1]+dp[i-k];
                dp[i]%=mod;
                sum[i]=sum[i-1]+dp[i];
                sum[i]%=mod;
            }
        }

        int a,b;
      for(int i=1;i<=t;i++)
      {
        cin>>a>>b;
        cout<<(sum[b]-sum[a-1]+mod)%mod<<endl;//由于取模后sum[b]的值可能会小于sum[a]的值，所以要加上mod
      }
    return 0;
}