牛客网暑期ACM多校训练营（第二场）A .run

本文链接：https://blog.youkuaiyun.com/Acer12138/article/details/81148304

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行走方式的数量，考虑到行走和跑步两种状态，并且跑步不能连续超过一秒。通过动态规划的方法给出了详细的解答过程。

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

思路：

1. dp[i][0]表示到达走着到达i dp[i][1]表示跑着到达i

ACcode:

#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>

using namespace std;

const int maxn = 1e5+5;
const int mod = 1000000007;

int dp[maxn][2];

int main(){

    int K,Q;
    while(cin>>Q>>K)
    {
        memset(dp,0,sizeof(dp));
        for (int i = 1;i<maxn;i++){
            if ( i<K ) {
                dp[i][0] = dp[i-1][0] + 1;
            } else {
                dp[i][0] = (dp[i-1][0] + dp[i-1][1] + 1) % mod;
                dp[i][1] = (dp[i-K][0] + 1) % mod;
            }
        }
        int l,r;
        while(Q--){
            scanf("%d %d",&l,&r);
            printf("%d\n",((dp[r][0] + dp[r][1]) % mod - (dp[l-1][0] + dp[l-1][1]) % mod + mod) % mod);
        }
    }
    return 0;
}