Codeforces 474D (五一训练 F)+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 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

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 lines contain two integers a_i and b_i (1 ≤ a_i ≤ b_i ≤ 10⁵), describing thei-th test.

Output

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).

Sample test(s)

Input

3 21 32 34 4

Output

655

Note

• 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). 

解析：一看就懂的一道DP

#include<iostream>#include<string.h>#include<stdio.h>#include<algorithm>using namespace std;const int maxn=100010;const int MOD=1000000007;int dp[maxn];  int a,b,t,k;void init(){    for(int i=1;i<k;i++)dp[i]=1;    dp[k]=2;    for(int i=k+1;i<maxn;i++)        dp[i]=(dp[i-1]+dp[i-k])%MOD;    for(int i=1;i<maxn;i++)        dp[i]=(dp[i-1]+dp[i])%MOD;}int main(){    cin>>t>>k;    init();    while(t--)    {        cin>>a>>b;        printf("%d\n",(dp[b]-dp[a-1]+MOD)%MOD);    }    return 0;}