zoukankan      html  css  js  c++  java
  • 牛客第二场A-run

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

    dp。dp[i][j]表示到i这一步是跑的(1)还是走的(0)。

    #include<iostream>
    #include<stdio.h>
    #include<string.h>
    using namespace std;
    long long dp[100005][2];
    long long sum[100005];
    int main()
    {
        long long   n,m;
          scanf("%lld%lld",&n,&m);
         memset(dp,0,sizeof dp);
            dp[0][0]=1;
            for(int i=1;i<=100000;i++)
            {
                dp[i][0]=(dp[i-1][0]+dp[i-1][1])%1000000007;
                if(i>=m)
                dp[i][1]=dp[i-m][0]%1000000007;
            }
            sum[0]=1;
            for(int i=1;i<=100000;i++)
            {
                sum[i]=sum[i-1]+dp[i][0]+dp[i][1];
                sum[i]=sum[i]%1000000007;
            }
        for(int i=0;i<n;i++)
        {
            long long q,w;
            scanf("%lld%lld",&q,&w);
          cout<<(sum[w]+1000000007-sum[q-1])%1000000007<<endl;
        }
        return 0;
    }
  • 相关阅读:
    SQL_TRACE与tkprof分析
    mysql学习之-三种安装方式与版本介绍
    1400
    输出二叉树中所有从根结点到叶子结点的路径
    [置顶] 处世悬镜之舍之
    Azkaban2配置过程
    [置顶] 处世悬镜之舍之 二
    UVALIVE 5893 计算几何+搜索
    Paxos算法 Paxos Made Simple
    Spring AOP 详解
  • 原文地址:https://www.cnblogs.com/2014slx/p/9364637.html
Copyright © 2011-2022 走看看