F - Flowers

最新推荐文章于 2020-03-10 23:39:53 发布

Yuric

最新推荐文章于 2020-03-10 23:39:53 发布

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分类专栏： ACM

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F - Flowers

Time Limit:1500MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

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Description

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 a_i and b_i (1 ≤ a_i ≤ b_i ≤ 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 a_i and b_i flowers at dinner modulo 1000000007 (10⁹ + 7).

Sample Input

Input

Output

6
5
5

Hint

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).

小结：

基本的想法是dp，虽然是一道水题，但是能秒杀我还是很开心，至少证明了我面对dp并不是束手无策的！

状态转移方程：dp[i]=dp[i-1]+dp[i-k],dp[i]代表的是当字符长度为i时，方法的种数。

然后我感觉唯一需要注意的是，在这里，我开始的时候将dp[0]初始化为1，为了满足状态转移方程的需要，这样提交的时候，将区间内的dp值累加，即为最后答案，然而很可惜，这样交了n次，n次不过，然后我就有点绝望了...

大概4分钟后，我想到了解决方案，将数据进行处理，使dp[n]储存从1到n的左右dp值的总和（这里的复杂度为n），然后要取区间[x,y]的方法数的时候，则ans=dp[y]-dp[x],

运行时间大大减小，于是就顺利过了，当然在进行相减的时候，需要将dp[0]赋值为0，为了避免特殊情况...

以下是AC代码：

#include<stdio.h>
#include<string.h>
#define maxn 100005
#define mod 1000000007
long long dp[maxn];
int main()
{
int T,k,max=0;
scanf("%d%d",&T,&k);
dp[0]=1;
for(int i=1;i<=100000;i++)
{
dp[i]=dp[i-1]+((i-k)>=0?dp[i-k]:0);
dp[i]%=mod;
}
for(int i=2;i<=100000;i++)
{
dp[i]+=dp[i-1];
}
dp[0]=0;
//for(int i=1;i<=4;i++)
//printf("%d%c",dp[i],i==4?'\n':' ');
for(int i=1;i<=T;i++)
{
int temp1,temp2;
scanf("%d%d",&temp1,&temp2);
printf("%lld\n",(dp[temp2]-dp[temp1-1])%mod);
}
return 0;
}