【CODEFORCES】 D. Flowers

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 modulo 1000000007 (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).
문제풀이: 이 문제는 비교적 간단한 동귀이다. d[i]를 설정하면 i송이 꽃을 먹었을 때의 총 방안수이다. 그러면 d[i]=d[i-1]+d[i-k]를 설정한다.주의해야 할 것은 d[0]=1, d[1]은 k=1에서 2와 같고 다른 때는 d[1]=1이다.이것 이외에 접두사와 구화로mod 출력 결과를 얻었습니다.
mod를 뽑는 것은 정말 기술적인 일이다.

#include <iostream>
#include <cstdio>
#include <cstdlib>

using namespace std;

int t,k,a[100005],b[100005],max1;
long long d[100005],ans,sum[100005];

int main()
{
    max1=0;
    scanf("%d%d",&t,&k);
    for (int i=0;i<t;i++)
    {
        scanf("%d%d",&a[i],&b[i]);
        max1=max(b[i],max1);
    }
    d[0]=1;
    if (k==1) d[1]=2;
    else d[1]=1;
    for (int i=2;i<k;i++) d[i]=d[i-1];
    for (int i=k;i<=max1;i++)
        d[i]=(d[i-1]+d[i-k])%1000000007;
    sum[0]=d[0];
    for (int i=1;i<=max1;i++)
        sum[i]=(sum[i-1]+d[i])%1000000007;
    for (int i=0;i<t;i++)
        printf("%I64d
",((sum[b[i]]+d[a[i]])%1000000007-sum[a[i]]%1000000007+1000000007)%1000000007);
    return 0;
}