题目链接:hdu5879 Cure 题解:用字符串输入.n很大时答案趋近与(π^2)/6. #include<cstdio> #include<algorithm> #include<iostream> #include<cstring> #include<queue> #include<cmath> using namespace std; ; const double pi = acos(-1.0); char a[N]; dou…

题目链接:http://acm.hdu.edu.cn/contests/contest_showproblem.php?pid=1002&cid=723 Cure Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7400    Accepted Submission(s): 1099 Problem Description Given a…

Cure Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1333    Accepted Submission(s): 440 Problem Description Given an integer n, we only want to know the sum of 1/k2 where k from 1 to n.   Input…

Cure Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 293    Accepted Submission(s): 96 Problem Description Given an integer n, we only want to know the sum of 1/k2 where k from 1 to n.   Input T…

论文 技术分析<关于网络分层信息泄漏点快速检测仿真> "1.基于动态阈值的泄露点快速检测方法,采样Mallat算法对网络分层信息的离散采样数据进行离散小波变换;利用滑动窗口对该尺度上的小波系数进行加窗处理,计算离散采样数据窗函数包含区间的小波熵,实现有效去噪和特征提取.2.将泄露点检测值和滑动窗口中平均熵值之间的差与动态阈值作比较,判断是否存在泄露点.""<基于云计算入侵检测数据集的内网用户异常行为分类算法研究>" "采用Weka机…

Cure Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4736    Accepted Submission(s): 1100 Problem Description Given an integer n, we only want to know the sum of 1/k2 where k from 1 to n.   Inpu…

A Cure for the Common Code Time Limit: 3000ms Memory Limit: 262144KB This problem will be judged on CodeForcesGym. Original ID: 100641B64-bit integer IO format: %I64d      Java class name: (Any)   You've been tasked with relaying coded messages to yo…

题目链接 题意:给定一个数n,求1到n中的每一项的平方分之一的累加和. 题解:题目没有给数据范围,而实际上n很大很大超过long long.因为题目只要求输出五位小数,我们发现当数大到一定程度时值是固定的 pi*pi/6.小的打表就行了,这里打表为了防止爆内存我用了优化的方法,类似于我之前写的light oj 1234. #include <iostream> #include <math.h> #include <stdio.h> #include <cstri…

链接:http://acm.hdu.edu.cn/showproblem.php?pid=5879 解法:我们知道到某个极限之后结果相同,所以找到那个极限,其他保存之后输出就好了 #include<stdio.h> //#include<bits/stdc++.h> #include<string.h> #include<iostream> #include<math.h> #include<sstream> #include<…

题目链接:传送门 题目大意:就是给四个数,和一个d,问1-d中有多少个数字不是那四个数的倍数; 这道题的d数据很小直接暴力可以过: 暴力代码:时间复杂度O(1): #include<stdio.h> int main(){ int k,m,n,d,l; scanf("%d%d%d%d%d",&k,&l,&m,&n,&d); ; ;i<=d;i++) || i%l== || i%n== || i%m==) ans++; print…