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 sizek

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 contains several test cases.

The first line contains two integers t andk (1?≤?t,?k?≤?10⁵), wheret represents the number of test cases.

The next t

Print t lines to the standard output. Thei-th line should contain the number of ways in which Marmot can eat betweena_i andb_i flowers at dinner modulo1000000007 (10⁹?+?7).

3 2
1 3
2 3
4 4

6
5
5

• For K = 2 and length1 Marmot can eat (R).
• For K = 2 and length2 Marmot can eat (RR) and (WW).
• For K = 2 and length3 Marmot can eat (RRR), (RWW) and (WWR).
• For K = 2 and length4 Marmot can eat, for example, (WWWW) or (RWWR), but for example he can‘t eat (WWWR).

考慮第n個。假如n是小於k的，那麽僅僅能都是是R，也就是僅僅有一種情況。

假如大於等於k。假設第n個是W，那麽從n-k+1到n所有為W，假設第n個是R，那麽數量就是前n-1個的數量。

dp[n] = 1; (0<= n < k)

dp[n] = dp[n-1] + dp[n-k]; (n >= k)

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <iostream>
#include <queue>
#include <algorithm>
#include <cmath>
#define mem(f) memset(f,0,sizeof(f))
#define M 100005
#define mod 1000000007
#define MAX 0X7FFFFFFF
#define maxn 100005
#define lson o<<1, l, m
#define rson o<<1|1, m+1, r
using namespace std;
typedef long long LL;

int n = maxn, k, t, a, b, dp[maxn], sum[maxn];

int main()
{
    scanf("%d%d", &t, &k);
    for(int i = 0; i < k; i++) dp[i] = 1;
    for(int i = k; i < n; i++) dp[i] = (dp[i-1] + dp[i-k])%mod;
    for(int i = 1; i < n; i++) sum[i] = (sum[i-1] + dp[i])%mod;
    while(t--) {
        scanf("%d%d", &a, &b);
        printf("%d\n", ((sum[b]-sum[a-1])%mod+mod)%mod );
    }
    return 0;
}

Codeforces Round #271 (Div. 2) D. Flowers （遞推 預處理）