題目描述

White Cloud is exercising in the playground.
White Cloud can walk 1 meters or run k meters per second.
Since White Cloud is tired,it can't run for two or more continuous seconds.
White Cloud will move L to R meters. It wants to know how many different ways there are to achieve its goal.
Two ways are different if and only if they move different meters or spend different seconds or in one second, one of them walks and the other runs.

輸入描述:

The first line of input contains 2 integers Q and k.Q is the number of queries.(Q<=100000,2<=k<=100000)
For the next Q lines,each line contains two integers L and R.(1<=L<=R<=100000)

輸出描述:

For each query,print a line which contains an integer,denoting the answer of the query modulo 1000000007.

示例1

輸入

複製

3 3
3 3
1 4
1 5

輸出

複製

2
7
11

#include<iostream>
#include<algorithm>
#include<string>
#include<map>//int dx[4]={0,0,-1,1};int dy[4]={-1,1,0,0};
#include<set>//int gcd(int a,int b){return b?gcd(b,a%b):a;}
#include<queue>
#include<cmath>
#include<stack>
#include<string.h>
#include<stdlib.h>
#include<cstdio>
#define ll long long
#define maxn 500005
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000") ///在c++中是防止暴棧用的
const int mod=((1e9)+7);
ll n,k,l,r;
ll dp[maxn][2];
ll sum[maxn];
/*
題目大意：一個人每秒可以走1一米，
也可選擇跑k米，但跑完後必須走不能連續跑，
給定l和r，問有多少中跑步方案跑l到r的距離。

首先方案數肯定是滿足字首和性質的，即
f(l,r)=sum(r)-sum(l-1);
那麼dp的思路就很清晰了，
dp[i][0]表示跑到i米最後一步是走的，
dp[i][1]表示跑到i米最後一次是跑的,
遞推性質很簡單，看程式碼即可，
然後計算個字首和即可。

*/
int main()
{
    scanf("%lld%lld",&n,&k);
    sum[0]=0;

    for(int i=1;i<k;i++)
    {
        dp[i][0]=1;
        sum[i]=sum[i-1]+1;
    }
    sum[k]=sum[k-1]+2;
    dp[k][1]=dp[k][0]=1;
    for(int i=k+1;i<maxn;i++)
    {
        dp[i][0]=(dp[i-1][0]+dp[i-1][1])%mod;
        dp[i][1]=(dp[i-k][0])%mod;
        sum[i]=(sum[i-1] + dp[i][0]+dp[i][1])%mod;
    }
    for(int i=1;i<=n;i++)
    {
        scanf("%lld%lld",&l,&r);
        printf("%lld\n",(sum[r]-sum[l-1]+mod)%mod);///////
    }
    return 0;
}