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 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.
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).
3 2 1 3 2 3 4 4
6 5 5
- 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：吃花朵，而且白花必须是连续吃k朵，问有多少种吃法
思路：推出状态转移方程，假设前一次满足条件的解为XXX，则如果当前吃的是红花，则部分最优解为XXXR，即dp[n-1]，如果当前吃的是白花，则必须是 XXX W..W（k个连续W）这样的情况，那么我们只需要知道前面XXX满足的条件即可，而结果就是dp[n-k]，所以状态转移为：dp[n]=dp[n-1]+dp[n-k]，接下来打表，存一下前缀和即可了。
#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<stack>
#include<queue>
#include<algorithm>
#include<string>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#define eps 1e-8
#define zero(x) (((x>0?(x):-(x))-eps)
#define mem(a,b) memset(a,b,sizeof(a))
#define memmax(a) memset(a,0x3f,sizeof(a))
#define pfn printf("\n")
#define ll __int64
#define ull unsigned long long
#define sf(a) scanf("%d",&a)
#define sf64(a) scanf("%I64d",&a)
#define sf264(a,b) scanf("%I64d%I64d",&a,&b)
#define sf364(a,b,c) scanf("%I64d%I64d%I64d",&a,&b,&c)
#define sf464(a,b,c,d) scanf("%I64d%I64d%I64d%I64d",&a,&b,&c,&d)
#define sf564(a,b,c,d,ee) scanf("%I64d%I64d%I64d%I64d%I64d",&a,&b,&c,&d,&ee)
#define sf2(a,b) scanf("%d%d",&a,&b)
#define sf3(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define sf4(a,b,c,d) scanf("%d%d%d%d",&a,&b,&c,&d)
#define sf5(a,b,c,d,ee) scanf("%d%d%d%d%d",&a,&b,&c,&d,&ee)
#define sff(a) scanf("%f",&a)
#define sfs(a) scanf("%s",a)
#define sfs2(a,b) scanf("%s%s",a,b)
#define sfs3(a,b,c) scanf("%s%s%s",a,b,c)
#define sfd(a) scanf("%lf",&a)
#define sfd2(a,b) scanf("%lf%lf",&a,&b)
#define sfd3(a,b,c) scanf("%lf%lf%lf",&a,&b,&c)
#define sfd4(a,b,c,d) scanf("%lf%lf%lf%lf",&a,&b,&c,&d)
#define sfc(a) scanf("%c",&a)
#define ull unsigned long long
#define pp pair<int,int>
#define debug printf("***\n")
#define pi 3.1415927
#define mod 1000000007
#define rep(i,a,b) for(int i=a;i<b;i++)
const double PI = acos(-1.0);
const double e = exp(1.0);
const int INF = 0x7fffffff;;
template<class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }
template<class T> T lcm(T a, T b) { return a / gcd(a, b) * b; }
template<class T> inline T Min(T a, T b) { return a < b ? a : b; }
template<class T> inline T Max(T a, T b) { return a > b ? a : b; }
bool cmpbig(int a, int b){ return a>b; }
bool cmpsmall(int a, int b){ return a<b; }
using namespace std;
int main()
{
//freopen("data.in","r",stdin);
//freopen("data.out" ,"w",stdout);
ll dp[100010];
ll flag[100010];
ll t,k;
while(~sf264(t,k))
{
mem(dp,0);
mem(flag,0);
dp[0]=1;
rep(i,1,1e5+1)
{
if(i>=k)
dp[i]=(dp[i-1]+dp[i-k])%mod;
else
dp[i]=dp[i-1]%mod;
dp[i]%=mod;
}
rep(i,1,1e5+1)
{
flag[i]=flag[i-1]+dp[i];
//flag[i]%=mod;
}
ll a,b;
while(t--)
{
sf264(a,b);
cout<<(flag[b]-flag[a-1])%mod<<endl;
}
}
return 0;
}