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
最近做数据分析,需要用到累加功能,发现强大的oracle还真有,用over(order by field) 例子: 数据表中最后一列就是累加的效果 累加sql: select t.acc_pedal_pos,count(*) num,sum(count(*)) over(order by t.acc_pedal_pos) accu_sum from GT1117CARDATA t where t.acc_pedal_pos>0 group by t.acc_pedal_pos order b
select s.slice_date_to ,s.made_id ,sum(s.steup_count)over(partition by s.made_id order by s.slice_date_to) from rpt_sliceresult_customer_setup s where s.group_type='SECOND_PRODUCE_DATE' and s.evaluate_id='CQM20150818112644734'; select s.slice_dat