补题：牛客网暑期ACM多校训练营（第二场）-A-run

本文链接：https://blog.youkuaiyun.com/Li_Hongcheng/article/details/81149608

本文介绍了一种通过动态规划解决健身路径计数问题的方法。给定行走和跑步的速度及允许的移动范围，算法计算出所有可能的运动组合方式。重点在于避免连续跑步，并确保最终移动的距离落在指定区间内。

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题目大意
白云在健身，每秒可以走1米或跑k米，并且不能连续两秒都在跑。
当它的移动距离在[L,R]之间时，可以选择结束锻炼。
问有多少种方案结束。

题目描述
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.

输入描述:
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)
输出描述:
For each query,print a line which contains an integer,denoting the answer of the query modulo 1000000007.
示例1
输入
复制
3 3
3 3
1 4
1 5
输出
复制
2
7
11

学习：

DP f[i][0/1]表示已经跑了i米，最后一步是跑还是走的方案数。
f[i][1]=f[i-k][0],f[i][0]=f[i-1][0]+f[i-1][1] 前缀和
sum[i]=(sum[i-1]+dp[i][1]+dp[i][0])%mod;

#include<bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
int main()
{
    int q,k;
    int l,r;
    int dp[100005][2];
    int sum[100005];
    while(~scanf("%d%d",&q,&k))
    {
        memset(sum,0,sizeof(sum));
        memset(dp,0,sizeof(dp));
        dp[0][0]=1;
        for(int i=1; i<=100005; i++)
        {
            dp[i][0]=(dp[i-1][0]+dp[i-1][1])%mod;
            if(i>=k)
            {
                dp[i][1]=dp[i-k][0];
            }
            sum[i]=(sum[i-1]+dp[i][1]+dp[i][0])%mod;
        }
        while(q--)
        {
            scanf("%d%d",&l,&r);
            cout<<(sum[r]-sum[l-1]+mod)%mod<<endl;
        }
    }
}