codeforces Round #271(div2) D解题报告

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).

题目大意：

给出一堆R字母和W字母，要求求出不同组合的个数，其中W字母的个数必须为K的倍数。

解法：

动态规划，利用之前的状态数。

首先，我们根据样例可以知道，第i长度的状态数，可以完全继承第i-1长度的状态数，因为对R无限制，只需要在所有的旁边加个R就够了，且不会有重复。

然而，当其长度i可以被k整处的时候，那么就意味着可以继承之前的第i-k长度的状态数，因为W可以再增加k个且连续。

总结一下状态方程式就是：

f[i]=f[i-1]+((i%k==0) ? 0:f[i-k]);

代码：

#include 

#define modn 1000000007

int t, k;
int ans[100010], sum[100010];

void init() {
	scanf("%d%d", &t, &k);

	ans[0]=1;
	for (int i = 1; i <= 100000; i++) {
		ans[i]=ans[i-1];

		if (i >= k) {
			ans[i] += ans[i-k];
			if (ans[i] >= modn)  ans[i]-=modn;
		}

		sum[i] = sum[i-1]+ans[i];
		if (sum[i] >= modn)  sum[i] -= modn;
	}
} 

void solve() {
	for (int i = 1; i <= t; i++) {
		int tmpa, tmpb;

		scanf("%d%d", &tmpa, &tmpb);
		printf("%d\n", (sum[tmpb]-sum[tmpa-1]+modn)%modn);
	}
}

int main() {
	init();
	solve();
}