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RSS订阅原创 java第几次作业?忘了
import javax.swing.*;import java.awt.*;import java.awt.event.*;import javax.swing.event.*;public class Main { public static void main(String[] arg) throws Exception{ final JFrame jfr...
2018-05-01 12:12:42
221
原创 矩阵迭代
#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<string>#include<vector>#include<stack>#include<bitset>#in
2018-05-01 08:56:35
2138
原创 计算器界面
import java.awt.*;import javax.swing.*;public class LabelFrame extends JFrame{ private final JButton []buttons; private static final String []names = {"7","8","9","/","4","5","6",&q
2018-04-24 19:33:44
272
原创 DIJISTRA
#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<string>#include<vector>#include<stack>#include<bitset>#in
2018-04-24 08:51:01
229
原创 矩阵计算
#include<iostream>#include<cmath>using namespace std;typedef long long ll;const double eps = 1e-6;const int maxn = 100;double a[maxn][maxn];double x[maxn],y[maxn],b[maxn];double l[...
2018-04-22 15:12:28
408
原创 牛顿法（作业用）
#include <cmath>#include <iostream>#include <stdio.h>const double eps = 1e-5;using namespace std;double newton1(){ double x0,x1; x0 = 0.5; double fx = x0*x0 - exp(x0); double fx1 =
2018-04-16 13:37:07
222
原创 牛顿迭代和弦截法
#include<cstdio>#include<algorithm>#include<iostream>#include<cmath>using namespace std;const double eps = 1e-5;double newton1(){ double x0,x1; x0 = 0.5; doub...
2018-04-10 10:58:56
2510
原创 霍夫曼编码简单实现（无链表）
#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<string>#include<vector>#include<stack>#include<bitset>#include<cstdlib>#include<cmath>#include<set>#include<lis
2018-04-10 09:05:22
1223
原创 java how to programme TestWindow
import java.awt.*;import java.awt.event.*;import javax.swing.*;// a [x2 +] b[x+]c [=0] public class TestWindow extends JFrame{ private JLabel alabel1,alabel2,alabel3; private JTextField aFi
2018-04-09 14:44:15
170
原创 java how to programme 异常类
package multiple;import java.util.Scanner;public class exceptionTest { static int a[]= new int[100]; public static void main (String[] args) { Scanner cin = new Scanner (System.in);
2018-04-09 09:23:01
174
原创 Java9.6
public class Shape { public static void main (String[] args) { shapes s1 = new shapes ("Red","polygon"); System.out.printf ("%s\n",s1.toString() ); TwoDimensionShape t2 = ne
2018-04-02 08:49:53
168
原创 java how to program 9.3
// UndergraduateStudent and GraduateStudent//Freshman, Sophomore, Junior and Senior might extend UndergraduateStudent, and DoctoralStudent and MastersStudent might be subclasses of GraduateStudent. Af
2018-04-02 08:22:32
523
原创 java how to program 8.15
package multiple;public class Rational { public int numerator,denominator; public int gcd (int a,int b) { if (b==0) return a; else return gcd (b,a%b); } public Rational
2018-04-01 10:57:14
279
原创 友元函数简单应用
#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<string>#include<vector>#include<stack>#include<bitset>#include<cstdlib>#include<cmath>#include<set>#include<lis
2018-03-29 20:05:20
241
原创 一份画图的java小作业
如题import java.awt.Color;import java.awt.Graphics;import javax.swing.JPanel;public class DrawRainbow extends JPanel{// define indigo and violet private final static Color VIOLET = new Color(128,
2018-03-27 20:16:37
212
原创 莫比乌斯反演一点笔记
首先是求逆元，exgcd的就不在这里说了ll get_inv(ll n,ll p){ inv[1]=1; for (int i=2;i<=n;i++){ inv[i]=(ll)(p-p/i)*inv[p%i]%p; }}貌似这个常用反推inv[300000]= powmod (c[300000],mod-2); for (ll i=300000-1;
2018-03-21 20:47:03
203
原创 析构构造函数以及xcode的断点
代码#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<string>#include<vector>#include<stack>#include<bitset>#include<cstdlib>#include<cmath>#include<set>#include<l
2018-03-18 11:11:06
239
原创 析构构造函数以及xcode的断点
代码#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<string>#include<vector>#include<stack>#include<bitset>#include<cstdlib>#include<cmath>#include<set>#include<l
2018-03-18 11:10:56
287
原创 E - Modified GCD CodeForces - 75C
水题#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<string>#include<vector>#include<stack>#include<bitset>#include<cstdlib>#include<cmath>#include<set>#include<l
2018-02-01 21:34:14
233
原创 GCD Table CodeForces - 583C
emm#include<stdio.h>#include<string.h>#include<queue>#include<map>#include<algorithm>using namespace std;int pre[505];int a[505*505];int gcd(int a,int b){ if(b==0)return a; else return
2018-02-01 21:32:18
204
原创 CodeForces - 798C Mike and gcd problem
emm纸老虎的构造题吧，题还是不错的#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<string>#include<vector>#include<stack>#include<bitset>#include<cstdlib>#include<cmath>#include
2018-02-01 21:30:01
210
原创 Maximal GCD CodeForces - 803C
emmm 求一个数n大于k的最小约数这样的问题，如果不暴力的话其实可以用欧拉筛法的思想遍历n的素因子，每个素因子求一个约数这样去弄，会比直接暴力省时间。#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<string>#include<vector>#include<stack>
2018-02-01 21:26:52
202
原创 codeforces672 A (数字转换成字符串)
把数字转换成字符串，一种写法to_string但编译器不一定支持。#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <string>#include <vector>#include <stack>#include <bitset>#include <cstdlib>
2017-11-06 19:59:26
634
原创 codeforces883I 2017-2018 ACM-ICPC, NEERC, Southern Subregional Contest (Online Mirror, ACM-ICPC Rule
这题还是不错的，题目条件很容易想到是二分答案去做，大于k的集合的划分为了满足一个最小化最大值，这是一个二分的条件。问题是怎么check，在下只想出了笨办法，n的遍历再更新线段树，所以总共nlogn91og10，肯定是可以做的，但是烦。能很快想出dp的做法也是本事。#include <cstdio>#include <cstring>#include <algorithm>#include <
2017-11-01 19:37:12
465
原创 HDU 5514 Frogs HDU 4135 Sum (容斥原理的应用)
昨天想出了frogs这道题的做法，通过等差数列减小求公因子加和之类的问题的时间复杂度，网上一搜发现是一种容斥原理的应用。今天随便找一道数论题目看看，没想到又发现一道类似的题，这次是求区间内与p互斥的数的加和，同样可以用这种性质数组做记录然后等差数列求和。
2017-10-26 22:19:11
214
原创 String Reconstruction CodeForces - 828C
在下还是太弱了#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <string>#include <vector>#include <stack>#include <bitset>#include <cstdlib>#include <cmath>#include
2017-10-12 21:38:41
307
原创 Hacker, pack your bags! CodeForces - 822C
#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <string>#include <vector>#include <stack>#include <bitset>#include <cstdlib>#include <cmath>#include <set>#
2017-10-11 10:36:08
415
原创 Codeforces 813C The Tag Game
理解题// 38#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <string>#include <vector>#include <stack>#include <bitset>#include <cstdlib>#include <cmath>#inclu
2017-09-28 16:22:21
233
原创 H - Partial Tree HDU - 5534 (2015年长春区域赛)
15年长春区域赛的题，属于能力范围的题，题目的模型很清楚，往dp角度想一想可以想出完全背包的模型。#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <string>#include <vector>#include <stack>#include <bitset>#i
2017-09-27 21:31:53
266
原创 Codeforces 812C Sagheer and Nubian Market
二分买的产品的数量，排序，最小cost，简单题。#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <string>#include <vector>#include <stack>#include <bitset>#include <cstdlib>#include
2017-09-26 15:09:56
313
原创 HDU1565方格取数
#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <string>#include <vector>#include <stack>#include <bitset>#include <cstdlib>#include <cmath>#include <set>#
2017-09-23 10:12:35
257
原创 POJ3281 Dining
经典的网络流建模题，开始其实就想到food连牛连drink，但没想到怎么控制一头牛只吃一份food和一份drink，后来换了一种想法是一头牛连一头左牛，一头右牛，分别连饮料，食物，这样可以做到把一头牛吃的食物和drink分开来，但还是不能解决一头牛各吃一份，最后想起开始的思路，发现一头牛可以拆成左牛右牛控制流量啊，bingo。#include <cstdio>#include <cstring>
2017-09-23 10:02:43
229
原创 HDU 4614 Vases and Flowers
2013年多校赛的题，还是很好的，明显的线段树，主要是一操作的处理，可以二分再查询，确定开始的和结束的位置。写起来比较麻烦，先占个坑，代码码完再贴。
2017-09-19 23:11:17
292
原创 HDU 1074
题意是要做n个作业，每个作业给出花的时间和deadline，超出deadline一天多扣一费，让你输出最小的超出费用，并输出写作业的顺序。易想到将用1
2017-09-14 07:53:20
195
原创 HDU 6070 Dirt Ratio
暑假多校赛的一道题，印象深刻，今天终于补了，现在感觉也不是特别难，题意是很经典的那种问题，就是给你一个数列，问一个区间不同的个数比区间的长度的值，在这个数列里的最小值。之前搜过区间不同数的个数的查询问题，发现了一个叫主席数的东西。（不会）那么这道题应该是更难了一点，求不同数的个数还要比上区间的长度，求一个最小值。这个时候看到这种分数很容易想到分数规划，这道题就可以列个式子。x/ (r-l+1) <=
2017-09-13 19:43:00
210
原创 HDU 6199 gems gems gems (2017沈阳网络赛)
比赛的时候想出了DP的递推式，感觉好像可以做但又不自信，后来又想错了dp[i][j],第二维其实只有200多。（又急着走结果就否定掉自己思路了）递推式类似dp1[i][j] = max (dp2[i-j][j],dp2[i-j][j-1]) + pre[i]-pre[j]这样，dp2同理，因为是第一个人先动，所以最后会是dp1的最优值。但是这道题内存卡的小，看了网上大神的题解，有通过指针，还有滚动
2017-09-11 12:32:22
276
原创 HDU 1079
写模拟什么的最累了，先写到这一个代码雏形贴这儿。#include <algorithm>#include <bitset>#include <cassert>#include <climits>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <deque>#include <
2017-09-09 20:01:57
273
原创 HDU 2897 邂逅明下
题意是给n个钱币，每个人最少取p颗，最多q颗，少于p颗时必须全部全完，取最后一下的人输，判断先手是赢还是输。完全就是那个取1——m的模型的类推，把1换成p，m换成q，于是用类似的做法解决。#include <algorithm>#include <bitset>#include <cassert>#include <climits>#include <cmath>#include <cst
2017-09-09 19:09:19
187
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