D. Flowers Codeforces Round #271(div2)

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).

状态转移方程 dp[i]=dp[i-1]+dp[i-k]。

取模要特别注意。

代码：

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
const long long mod = 1000000007;
const int maxn=100000;
long long dp[maxn+100];
long long sum[maxn+100];
using namespace std;
int main(){
    int n,k;
    memset(sum, 0, sizeof(sum));
    scanf("%d %d",&n,&k);
    sum[0]=0;
    dp[0]=1;
    for(int i=1;i<k;++i){
        dp[i]=1;
        sum[i]=sum[i-1]+dp[i];
    }
    for(int i=k;i<=maxn;++i){
        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=0;i<n;++i){
        scanf("%d %d",&a,&b);
        printf("%I64d\n",(sum[b]-sum[a-1]+mod)%mod);
    }
    return 0;
}