牛客网暑期ACM多校训练营（第二场）_A run_牛客网acm多校训练营第二场run用递归的方法-优快云博客

本文解析了一道来自牛客网的算法竞赛题，题目要求计算从起点到终点的不同跑步方式的数量，考虑到行走和跑步的特定规则。通过使用动态规划的方法解决了该问题，并提供了完整的C++代码实现。

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

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

输出

复制

2
7
11

白云每秒走1m 或者每秒跑 k米不能连续跑哦；

问白云跑到 L-R 之间有几种方式

= =dp 不知道怎么写活活搞成了递推。hhhh 反正也是过了

这是多校跟队友做出来的第一道题纪念一下嘻嘻

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
 
#define maxn (int)1e5+10
const int Mod=1e9+7 ;
long long int n,dp[maxn],c[50],m,a[maxn]={0},r;
void dabiao(int k)
{
    memset(dp,0,sizeof(dp));
    int i=1;
    dp[0]=1;
    while(i<k)
    {
            dp[i]=1;
            ++i;
    }
    dp[k]=2;
    for(int i=k+1;i<=100000;i++)
    {
        dp[i]=(dp[i-k-1]+dp[i-1])%Mod;
        //cout<<dp[i]<<endl;
    }
    a[1]=1;
    for(int i=2;i<=100000;i++)
    {
        a[i]=(a[i-1]+dp[i])%Mod;
    //  cout<<a[i]<<endl;
    }
}
int main()
{
    int q,k,l,r;
    cin>>q>>k;
    dabiao(k);
    while(q--)
    {
        int ans=0;
        cin>>l>>r;
        int i=1;
        ans=((a[r]%Mod-a[l-1]%Mod)%Mod+Mod)%Mod;
        cout<<ans<<endl;
    }
    return 0;
}

。