语法格式：row_number() over(partition by 分组列 order by 排序列 desc)

row_number() over()分组排序功能：

在使用 row_number() over()函数时候，over()里头的分组以及排序的执行晚于 where 、group by、  order by 的执行。

例一：

表数据：

create table TEST_ROW_NUMBER_OVER(
       id varchar(10) not null,
       name varchar(10) null,
       age varchar(10) null,
       salary int null
);
select * from TEST_ROW_NUMBER_OVER t;

insert into TEST_ROW_NUMBER_OVER(id,name,age,salary) values(1,'a',10,8000);
insert into TEST_ROW_NUMBER_OVER(id,name,age,salary) values(1,'a2',11,6500);
insert into TEST_ROW_NUMBER_OVER(id,name,age,salary) values(2,'b',12,13000);
insert into TEST_ROW_NUMBER_OVER(id,name,age,salary) values(2,'b2',13,4500);
insert into TEST_ROW_NUMBER_OVER(id,name,age,salary) values(3,'c',14,3000);
insert into TEST_ROW_NUMBER_OVER(id,name,age,salary) values(3,'c2',15,20000);
insert into TEST_ROW_NUMBER_OVER(id,name,age,salary) values(4,'d',16,30000);
insert into TEST_ROW_NUMBER_OVER(id,name,age,salary) values(5,'d2',17,1800);


一次排序：对查询结果进行排序（无分组）

select id,name,age,salary,row_number()over(order by salary desc) rn
from TEST_ROW_NUMBER_OVER t

结果：



进一步排序：根据id分组排序

select id,name,age,salary,row_number()over(partition by id order by salary desc) rank
from TEST_ROW_NUMBER_OVER t

结果：



 再一次排序：找出每一组中序号为一的数据

select * from(select id,name,age,salary,row_number()over(partition by id order by salary desc) rank
from TEST_ROW_NUMBER_OVER t)
where rank <2

结果：



排序找出年龄在13岁到16岁数据，按salary排序

select id,name,age,salary,row_number()over(order by salary desc)  rank
from TEST_ROW_NUMBER_OVER t where age between '13' and '16'

结果：结果中 rank 的序号，其实就表明了 over(order by salary desc) 是在where age between and 后执行的



例二：

1.使用row_number（）函数进行编号，如

select email,customerID, ROW_NUMBER() over(order by psd) as rows from QT_Customer

原理：先按psd进行排序，排序完后，给每条数据进行编号。

2.在订单中按价格的升序进行排序，并给每条记录进行排序代码如下：

select DID,customerID,totalPrice,ROW_NUMBER() over(order by totalPrice) as rows from OP_Order

3.统计出每一个各户的所有订单并按每一个客户下的订单的金额 升序排序，同时给每一个客户的订单进行编号。这样就知道每个客户下几单了：

select ROW_NUMBER() over(partition by customerID  order by totalPrice)
 as rows,customerID,totalPrice, DID from OP_Order

4.统计每一个客户最近下的订单是第几次下的订单：

with tabs as  
(  
select ROW_NUMBER() over(partition by customerID  order by totalPrice)
 as rows,customerID,totalPrice, DID from OP_Order  
 )  
select MAX(rows) as '下单次数',customerID from tabs 
group by customerID 

5.统计每一个客户所有的订单中购买的金额最小，而且并统计改订单中，客户是第几次购买的：

思路：利用临时表来执行这一操作。

1.先按客户进行分组，然后按客户的下单的时间进行排序，并进行编号。

2.然后利用子查询查找出每一个客户购买时的最小价格。

3.根据查找出每一个客户的最小价格来查找相应的记录。

    with tabs as  
     (  
    select ROW_NUMBER() over(partition by customerID  order by insDT) 
as rows,customerID,totalPrice, DID from OP_Order  
    )  
     select * from tabs  
    where totalPrice in   
    (  
    select MIN(totalPrice)from tabs group by customerID  
     ) 

6.筛选出客户第一次下的订单。

思路。利用rows=1来查询客户第一次下的订单记录。

    with tabs as  
    (  
    select ROW_NUMBER() over(partition by customerID  order by insDT) as rows,* from OP_Order  
    )  
    select * from tabs where rows = 1 
    select * from OP_Order 

7.注意：在使用over等开窗函数时，over里头的分组及排序的执行晚于“where，group by，order by”的执行。

    select   
    ROW_NUMBER() over(partition by customerID  order by insDT) as rows,  
    customerID,totalPrice, DID  
    from OP_Order where insDT>'2011-07-22' 

 

antd+精度状态控制，会出现第一次点击聚焦仍然显示上一次值的情况。
1、首先需要inputnumber对精度进行控制，这里精度控制在1-5
 <Form.Item labelCol={{ span: 8 }} wrapperCol={{ span: 8 }} label={"流速小数位数"}>
                    {form.getFieldDecorator('flowrateDecimallen', {
                      initialValue: objectData?.flowrateDecimallen,
                      rules: [
                        { validator: (rule: any, value: any, callback: any) => validateHeightAndMin(value, callback, '') }
                      ]
                    })(<InputNumber min={0} max={5} precision={0} placeholder={"请输入"} onChange={currentPrecisionChange}></InputNumber>)}
                  </Form.Item>

2、受精度控制的inputnumber
初始化数据：
  const initDymAttr = () => {
    dymAttr.flowrateInit = undefined
    dymAttr.flowInit = undefined
    dymAttr.cumulativeFlowMax = undefined
    dymAttr.cumulativeFlowMin = undefined
    dymAttr.currentSpeedMax = undefined
    dymAttr.currentSpeedMin = undefined
  }

状态：
 const [currentPrecision, setCurrentPrecision] = useState<number | undefined>(5);

 <Form.Item labelCol={{ span: 8 }} wrapperCol={{ span: 8 }} label={"流速初值"}>
                    {form.getFieldDecorator('flowrateInit', {
                      initialValue: objectData?.flowrateInit,
                      rules: [
                        { validator: (rule: any, value: any, callback: any) => validateHeightAndMin(value, callback, '') }
                      ]
                    })(<InputNumber onChange={(value: any) => keyChange(value, "flowrateInit", "current")}
                      onFocus={currentNumberFocus} precision={currentPrecision} placeholder={"请输入"}></InputNumber>)}
                  </Form.Item>

3、精度值变化响应函数：
  const currentPrecisionChange = (value: number | undefined) => {
    if (value != undefined && (value > 5 || value < 0)) {
      message.error("小数位为0-5位")
    }
    else {
      setCurrentPrecision(value)
    }
  }

4、受控组件值变化响应函数：
  const keyChange = (value: any, attrName: string, attrType: string) => {
    if (dymAttr[attrName] != value) {
      dymAttr[attrName] = value
      if (dymAttr[attrName] != null && parseFloat(dymAttr[attrName]).toFixed(currentPrecision) != "NaN" && attrType == "current") {
        form.setFieldsValue({ [attrName]: parseFloat(dymAttr[attrName]).toFixed(currentPrecision) })
        const textDom: any = document.getElementById(attrName);
        textDom.focus()
      }
      else if (dymAttr[attrName] != null && parseFloat(dymAttr[attrName]).toFixed(flowPrecision) != "NaN" && attrType == "flow") {
        form.setFieldsValue({ [attrName]: parseFloat(dymAttr[attrName]).toFixed(flowPrecision) })
        const textDom: any = document.getElementById(attrName);
        textDom.focus()
      }
    }
  }

5、受控组件聚焦响应函数：
  const currentNumberFocus = (e: any) => {
    var fieldName = e.target?.id
    if (dymAttr[fieldName] != null && parseFloat(dymAttr[fieldName]).toFixed(currentPrecision) != "NaN") {
      form.setFieldsValue({ [fieldName]: parseFloat(dymAttr[fieldName]).toFixed(currentPrecision) })
    }
  }

纠结了一段时间都没找出方法，最后灵光一现想出这个方法，没想到测试下果然成功了！

看目前网上几乎很难找到相对应的解决方法，所以这里分享出来，如果有更佳方法欢迎提出。
方法如下：
<input type="number" step="0.01" />

step 属性规定输入字段的合法数字间隔
在input标签中添加step属性，此例子中将step设为"0.01"，即可以输入含两位小数的数字。(仅输入含一位小数或整数也可)

可以自行修改step值来修改精确度。

Volodymyr’s favourite number is A and it has an odd number of positive divisors. When you add K to this number, the resulting sum also has an odd number of positive divisors. Given K, find all possible values of A

Volodymyr’s favourite number.

Input

The only line of input contains K a positive integer not exceeding 1e9
Output

On the first line, print the number of possible values of A. If the number of such values is greater than zero, on the second line print all possible values of A  in ascending order, separated by a single space.

Examples

Input
1

Output

0

Input

2

Output

1
-1

Note
Note that since zero has infinitely many divisors, it cannot be classified as number with an odd number of divisors.

题意很简单，给你一个数K,让你找一个A，A要满足A有奇数个正因子，且A加K后仍有奇数个正因子。

思路：首先你要知道只有平方数才有奇数个因子，所以，A要为平方数或者平方数的负数形式。且得到的也为平方数或者平方数的负数形式。

1.(a^2 )+ (b ^2)=k 压入aa,bb

2.(a^2 )-k =(b ^2) 压入-aa,bb

``      记得去除零的情形
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
set<ll>s;
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	ll k;
	cin>>k;
	ll a,b,a1,b1;
	for(int i=1;i*i<=k;i++)
	{
		a1=k-i*i;
		a=sqrt(a1);
		if(a*a==a1)
		{
			if(i*i!=0&&a!=0)
			{
				s.insert(-i*i);
				s.insert(-a1);	
			}					
		}
		
		if((k-i*i)%(2*i)==0)
		{
			a=(k-i*i)/(2*i);
			a1=a*a;
			b1=k+a1;
			if(a1==0||b1==0)
			{
				continue;
			} 
			s.insert(a1);
			s.insert(-b1);
		}
		
	}
	cout<<s.size()<<endl;
	set<ll>::iterator it=s.begin();
	for(;it!=s.end();it++)
	{
		cout<<*it<<" ";
	}
	
}