最近项目碰到一个新的需求,统计每日充值/消费之后的余额.对于这种需求,其实也很简单,只需要在每次充值/消费后,计算下余额,然后保存下来就可以了.但是对于这种需求,一条sql就能搞定,都不需要做冗余字段. 用图表展示会更详细: 要求的结果: MySQL写法一: select t.* ,(select sum(price) from t_charge temp where temp.date <= t.date) as total_price from t_charge t group by t.i
Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 10711 Accepted: 3182 Description Farmer John's N cows (1 ≤ N ≤ 100,000) share many similarities. In fact, FJ has been able to narrow down the list of features shared by his cows to a list
[题解] 二重循环枚举起始列和终止列,竖着往下加,转化为一个最大子段和问题,逐行累加即可. #include<cstdio> #include<cstring> #include<algorithm> #define LL long long #define N 1000 #define rg register using namespace std; int n,m,x; LL ans,sum[N][N]; inline int read(){ ,f=; char c
自己做的一个小项目里需要用mysql计算一些信息. mysql中的 表如下(注:表中数据都是测试数据,随机生成的) mysql> select * from shake_log; +-----+-------------+--------+---------------------+------+ | id | mobile | credit | time | type | +-----+-------------+--------+---------------------+------+ |