zoukankan      html  css  js  c++  java
  • Codeforces Round #337 (Div. 2) A水

    A. Pasha and Stick
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Pasha has a wooden stick of some positive integer length n. He wants to perform exactly three cuts to get four parts of the stick. Each part must have some positive integer length and the sum of these lengths will obviously be n.

    Pasha likes rectangles but hates squares, so he wonders, how many ways are there to split a stick into four parts so that it's possible to form a rectangle using these parts, but is impossible to form a square.

    Your task is to help Pasha and count the number of such ways. Two ways to cut the stick are considered distinct if there exists some integer x, such that the number of parts of length x in the first way differ from the number of parts of length x in the second way.

    Input

    The first line of the input contains a positive integer n (1 ≤ n ≤ 2·109) — the length of Pasha's stick.

    Output

    The output should contain a single integer — the number of ways to split Pasha's stick into four parts of positive integer length so that it's possible to make a rectangle by connecting the ends of these parts, but is impossible to form a square.

    Sample test(s)
    Input
    6
    Output
    1
    Input
    20
    Output
    4
    Note

    There is only one way to divide the stick in the first sample {1, 1, 2, 2}.

    Four ways to divide the stick in the second sample are {1, 1, 9, 9}, {2, 2, 8, 8}, {3, 3, 7, 7} and {4, 4, 6, 6}. Note that {5, 5, 5, 5} doesn't work.

    应该是今年最后一次做cf了 时间很良心 但是自己确实水  只做出a woc 以后打cf 不能再求援于任何人 必须靠自己

    len 为周长 求有多少种组成长方形但不是正方形的组合

    !!!

    #include<bits/stdc++.h>
    using namespace std;
    #define LL __int64
    int main()
    {
        LL len;
        scanf("%I64d",&len);
        if(len<4||len%2==1)
        printf("0
    ");
        else
        {
        len=len/2;
        if(len%2==0)
        printf("%I64d
    ",len/2-1);
        else
            printf("%I64d
    ",len/2);
        }
        return 0;
    }
    

      

  • 相关阅读:
    java后台读取配置文件
    冒泡排序
    均分火柴
    Dos 批处理 Shutdown
    时间复杂度分析
    Python冒泡排序
    Python装饰器
    获取状态栏高度
    利用zxing生成二维码
    Android下利用zxing类库实现扫一扫
  • 原文地址:https://www.cnblogs.com/hsd-/p/5081158.html
Copyright © 2011-2022 走看看