28的顺数和易逆数也是多少

来源：蜘蛛抓取(WebSpider) 时间：2018-07-21 05:54 标签： 易逆数也
数往者顺知来者逆怎么理解
数往者顺知来者逆怎么理解
09-08-04 &匿名提问
夫子在《说卦传》里面提到这句话：“数往者顺，知来者逆”；由于《易经》及与《易经》相关的书对大众一直充满着神秘色彩，所以对它的解释可能很多，也很玄，从而忽视其中的一句话：“百姓日用而不知”的简单道理，正是《易经》一直充满着神秘，所以在民间被“高推圣境”了，其实被誉为群经之首的《易经》并没有讲我们所认为的玄学。
请登录后再发表评论!
这是易经里的话，解释有好多种，最好还是找精通易经的高人吧!百度里说的都很玄……
请登录后再发表评论!
数往者顺知来者逆一个六爻卦，上卦是往，下卦是来。如果要预测未来要顺推，从初爻到上六爻；如果要预测过去，也就是知来就要逆推，也就是从六爻推到初爻。
请登录后再发表评论!博客分类：
Time Limit:
MS (Java/Others)
Memory Limit:
K (Java/Others)Total Submission(s): 1143
Accepted Submission(s): 387
Problem Description
N(3&=N&=20000) ping pong players live along a west-east street(consider the street as a line segment). Each player has a unique skill rank. To improve their skill rank, they often compete with each other. If two players want to compete, they must choose a referee among other ping pong players and hold the game in the referee's house. For some reason, the contestants can’t choose a referee whose skill rank is higher or lower than both of theirs.The contestants have to walk to the referee’s house, and because they are lazy, they want to make their total walking distance no more than the distance between their houses. Of course all players live in different houses and the position of their houses are all different. If the referee or any of the two contestants is different, we call two games different. Now is the problem: how many different games can be held in this ping pong street?
The first line of the input contains an integer T(1&=T&=20), indicating the number of test cases, followed by T lines each of which describes a test case.Every test case consists of N + 1 integers. The first integer is N, the number of players. Then N distinct integers a1, a2 … aN follow, indicating the skill rank of each player, in the order of west to east. (1 &= ai &= 100000, i = 1 … N).
For each test case, output a single line contains an integer, the total number of different games.
Sample Input
Sample Output
题目大意：每一个人都有一个实力值，顺序就是他的位置，要求寻找一个对手，然后再寻找一个裁判，要求裁判的实力再他们之间，而且位置要在他们两人的中间，这样可以组成一场比赛。问总共可以组织多少场比赛？
解题思路：以自己为裁判，然后找左边有多少人比较小，再找右边有多少人比自己大，然后两个数相乘就是以他为裁判的比赛场数，当然还有相反的，找右边比自己小的，左边比自己大的，再相乘相加。这样子就变成了：和找逆序数、顺序数差不多了，找左边（右边）比自己大（小）有多少个数，用树状数组最好最快！所以两个for循环就全部找到，然后相乘相加，搞定！！！！！！！
题目链接：
我的代码：
#include &iostream&
#include &stdio.h&
#include &memory.h&
const int MAX = 100005;
int a[MAX+1], b[MAX+1];
int c[MAX+1], d[MAX+1], e[MAX+1], f[MAX+1];
int lowbit(int i)
return i&(-i);
void update(int i, int x)
while(i &= MAX)
i += lowbit(i);
int sum(int i)
int sum = 0;
while(i & 0)
sum += a[i];
i -= lowbit(i);
int main()
scanf("%d", &t);
while(t--)
scanf("%d", &n);
memset(a, 0, sizeof(a));
for(i = 1; i &= i++)
scanf("%d", &b[i]);
for(i = 1; i &= i++)
update(b[i], 1);
c[i] = sum(MAX)-sum(b[i]);
//求左边有多少个数比自己大
d[i] = sum(b[i]-1);
//求左边有多少个数比自己小
memset(a, 0, sizeof(a));
for(i = i &= 1; i--)
update(b[i], 1);
e[i] = sum(MAX)-sum(b[i]);
//求右边有多少个数比自己大
f[i] = sum(b[i]-1);
//求右边有多少个数比自己小
for(i = 1; i &= i++)
///1.左边比自己大的数*右边比自己小的数
///2.左边比自己小的数*右边比自己大的数
ans += c[i]*f[i] + d[i]*e[i];
//相加就是总和
printf("%I64d\n", ans);
gzhu_101majia
浏览: 144377 次
来自: 广州
我特么顶上！！
基德KID.1412 写道
(window.slotbydup=window.slotbydup || []).push({
id: '4773203',
container: s,
size: '200,200',
display: 'inlay-fix'(答案是simon)
第三方登录：||Copyright & 2012 . All Rights Reserved .
网站制作：已解决问题
芯片的引脚顺序是怎么定义的,说清楚引脚朝上还是朝下,顺时针数还是逆时针数
提问时间： 11:29:16
浏览次数：6787
该答案已经被保护
普通芯片引脚图都俯视图即芯片面向面向看图 芯片般面1脚标记圆点整芯片起始脚边标记数据手册能知道 芯片数据手册图片俯视图定图片附件进行特殊标示、说明 按照俯视图四周管脚贴片型封装按照逆针向顺序递增管脚号。引脚朝逆针比 图& class=&ikqb_img_alink&&。
答案创立者
以企业身份回答&
快速解决你的电商难题
店铺优化排查提升2倍流量
擅长&nbsp 店铺优化
您可能有同感的问题

我要回帖

更多关于 易逆数也 的文章

 

随机推荐