zoukankan      html  css  js  c++  java
  • codeforces 363B

    B. Fence
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    There is a fence in front of Polycarpus's home. The fence consists of n planks of the same width which go one after another from left to right. The height of the i-th plank is hi meters, distinct planks can have distinct heights.

    Fence for n = 7 and h = [1, 2, 6, 1, 1, 7, 1]

    Polycarpus has bought a posh piano and is thinking about how to get it into the house. In order to carry out his plan, he needs to take exactly k consecutive planks from the fence. Higher planks are harder to tear off the fence, so Polycarpus wants to find such k consecutive planks that the sum of their heights is minimal possible.

    Write the program that finds the indexes of k consecutive planks with minimal total height. Pay attention, the fence is not around Polycarpus's home, it is in front of home (in other words, the fence isn't cyclic).

    Input

    The first line of the input contains integers n and k (1 ≤ n ≤ 1.5·105, 1 ≤ k ≤ n) — the number of planks in the fence and the width of the hole for the piano. The second line contains the sequence of integers h1, h2, ..., hn (1 ≤ hi ≤ 100), where hi is the height of the i-th plank of the fence.

    Output

    Print such integer j that the sum of the heights of planks jj + 1, ..., j + k - 1 is the minimum possible. If there are multiple such j's, print any of them.

    Examples
    input
    7 3
    1 2 6 1 1 7 1
    output
    3
    Note

    In the sample, your task is to find three consecutive planks with the minimum sum of heights. In the given case three planks with indexes 3, 4 and 5 have the required attribute, their total height is 8.

    题目的意思是输入栅栏总数n和需要拔的栅栏数k,然后输入n个数,计算连续拔的k个栅栏的最小长度,输出最小长度的开始栅栏。

    #include<iostream>
    #define MAXN 500005
    using namespace std;
    int fence[MAXN];
    int minfence[MAXN];
    int main()
    {
        int n, k, i, j=0;
        memset(minfence, 0, MAXN);
        cin >> n >> k;
        for (i = 0;i < n;++i)
        {
            cin >> fence[i];
            if (i < k)
                minfence[0] += fence[i];
        }
        int min = minfence[0];
        for (i = 1;i < n - k + 1;++i)
        {
            minfence[i] = minfence[i - 1] + fence[k + i - 1] - fence[i - 1];
            if (min > minfence[i])
            {
                min = minfence[i];
                j = i;
            }
        }
        cout << j+1 << endl;
        return 0;
    }
  • 相关阅读:
    .gitignore不生效-git上传忽略解决方案的配置文件
    JAVA
    Java
    python-函数基础01
    Java常见数据结构
    分代垃圾回收机制及垃圾回收算法
    设计模式【9】------>观察者模式
    设计模式【8】------>策略模式
    设计模式【7】------>原型模式
    设计模式【6】------>外观模式
  • 原文地址:https://www.cnblogs.com/chenruijiang/p/8289811.html
Copyright © 2011-2022 走看看