Flowers(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 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).
 
제목:
t와 k를 주면 t팀의 문의가 있음을 의미합니다.k는 한 번에 K송이 W꽃만 먹을 수 있음을 의미하며, 이후 t조의 a-b 구간을 제시한다.이 구간 내의 길이 서열이 조건을 충족시키는 방법수를 물어본다.
 
아이디어:
      DP.dp[i]=dp[i-k]+dp[i-1]는 한 번에 K송이 W꽃을 먹거나 R꽃을 한 번에 한 송이 먹는 것을 의미한다.마지막 답안의 뺄셈을 기억하고 + MOD +% MOD를 넣으세요.안 그러면 실수할 거야.
 
      AC:

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long ll;

const ll MOD = 1000000007;
const int MAX = 100005;

ll dp[MAX];
ll sum[MAX];
ll a[MAX], b[MAX];

int main() {

    int t, k;
    scanf("%d%d", &t, &k);

    ll Max = 0;
    for (int i = 1; i <= t; ++i) {
        scanf("%I64d%I64d", &a[i], &b[i]);
        Max = max(Max, b[i]);
    }

    for (int i = 0; i < k; ++i) {
        dp[i] = 1;
        sum[i] = sum[i - 1] + dp[i];
    }

    for (int i = k; i <= Max; ++i) {
        dp[i] = (dp[i - k] + dp[i - 1]) % MOD;
        sum[i] = (dp[i] % MOD + sum[i - 1] % MOD) % MOD;
    }

    for (int i = 1; i <= t; ++i) {
        printf("%I64d
", (sum[b[i]] - sum[a[i] - 1] + MOD) % MOD);
    }

    return 0;
}