（第二场）A Run 【动态规划】

链接：https://www.nowcoder.com/acm/contest/140/A

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.

案例

输入：

3 3
3 3
1 4
1 5

输出：

2
7
11

大概题意：

小白喜欢运动，他每秒可以走一米或者跑K米，有Q次查询，每次查询求跑0 ~ 【L，R】米的方案数；

思路：

一开始看到这道题想着是道规律题，结果花时间列数据找规律。。。

果然还是too young too simple，师兄花了短短的时间AC之后在群里丢给我们八个字“记忆化搜索，过了”。

想了那么多，这就是道纯粹的DP了啊，关键在于分开跑和走的方案，不能连续跑

状态：

dp[i][0] 走 i  meters 的方案数

dp[i][1] 跑 i meters 的方案数

转移方程：

dp[i][0] = dp[i-1][1] + dp[i-1][0];

dp[i][1] = dp[i-k][0] (注意题目条件：不能连续跑，有队友一直卡在60%这里的原因）

!!!更新dp值的同时也要更新前缀和sum[i] (用于最后求答， 本人傻傻的还在后面查询的时候才计算前缀和，一直T）!!!

AC code:

#include <cstdio>
#include <iostream>
#include <cstring>
#define mod 1000000007
using namespace std;

const int maxk = ;

int Q, k;
long long int dp[maxk][], sum[maxk];

int main()
{
    scanf("%d%d", &Q, &k);
    for(int i = ; i < k; i++)
    {
        dp[i][] = ;
        dp[i][] = ;
        if(i > )
        sum[i] = sum[i-] + dp[i][]+dp[i][];
    }
    dp[k][] = ;
    sum[k] = sum[k-] + dp[k][];
    for(int i = k; i < maxk; i++)
    {
        dp[i][] = ((dp[i-][] + dp[i-][]))%mod ;
        dp[i][] = dp[i-k][];
        sum[i] = sum[i-]+dp[i][]+dp[i][];
    }
    while(Q--)
    {
        int L, R;
        long long int ans = ;
        scanf("%d%d", &L, &R);
        ans =(sum[R] - sum[L-] + mod)%mod;
        printf("%lld\n", ans);
    }
    return ;
}