Partial Sum Accepted : Submit : Time Limit : MS Memory Limit : KB Partial Sum Bobo has a integer sequence a1,a2,…,an of length n. Each time, he selects two ends ≤l<r≤n and add |∑rj=l+1aj|−C into a counter which is zero initially. He repeats the selec
A. Flipping Game time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Iahub got bored, so he invented a game to be played on paper. He writes n integers a1, a2, ..., an. Each of those intege
这就是数学中的 A m n 的选取. 共有 m!/n!种可能.. 同样举一个例子吧.. 从12345这五个数字中随机选取3个数字,要求选出来的这三个数字是有序,也就是说从12345中选出来的是123这三个数的话,那么就有 123,132,312,321,213,231 这六种可能.. 好了.废话不多说了,上程序,解释写在城市的注释里. //A53 //排序,12345找出所有的排序组合 public class Test7 { static char[] ch; static String
[本文出自天外归云的博客园] 题1:求m以内的素数(m>2) def find_all_primes_in(m): def prime(num): for i in range(2, num): if divmod(num, i)[1] == 0: return False return True print([i for i in range(2, m + 1) if prime(i)]) if __name__ == '__main__': find_all_primes_in(100) 我
B. Soldier and Badges time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Colonel has n badges. He wants to give one badge to every of his n soldiers. Each badge has acoolness factor, which sh
/** * 设计4个线程,其中2个对num进行加操作,另两个对num进行减操作 */ public class ThreadTest { private int j; public static void main(String args[]) { ThreadTest tt = new ThreadTest(); Inc inc = tt.new Inc(); Dec dec = tt.new Dec(); for (int i = 0; i < 2; i++) { Thread t = ne
话说有一文章表article,存储文章的添加文章的时间是add_time字段,该字段为int(5)类型的,现需要查询今天添加的文章总数并且按照时间从大到小排序,则查询语句如下: 1 select * from `article` where date_format(from_UNIXTIME(`add_time`),'%Y-%m-%d') = date_format(now(),'%Y-%m-%d');或者: 1 select * from `article` where to_da
#include<cstdio> int f1(int a,int b) //最大公约数 { ) return b; else return f1(b,a%b); } int f2(int a,int b) //最小公倍数 { int g; g=a*b/f1(a,b); return g; } int main() { int t,i; scanf("%d",&t); while(t--) { int a,b; scanf("%d%d",&
python中, 实现列表中的整型元素两两相乘或列表中的数组元素两两相与 1. 假设列表中的元素是整型, 可调用以下函数: def list_any_two_mul(mylist): num = 1 temp = [] for i in mylist[:-1]: temp.append([i * j for j in mylist[num:]]) num = num + 1 # 把多个列表变成只有一个列表 re