C#格式化输出(记录)
int a = 12345678; //格式为sring输出 Label1.Text = string.Format("asdfadsf{0}adsfasdf",a); Label2.Text = "asdfadsf"+a.ToString()+"adsfasdf"; Label1.Text = string.Format("asdfadsf{0:C}adsfasdf"
USACO 2008 March Gold Cow Jogging 解题报告
题目链接:http://202.120.106.94/onlinejudge/problemshow.php?pro_id=143 这道题嘛,
POJ PKU 3659 Cell Phone Network 解题报告
题目链接:http://acm.pku.edu.cn/JudgeOnline/problem?id=3659 这题不算难题了,基本算是中等题 题
浙江理工 省赛总结 team62 By OWenT of Coeus
这次比赛成绩比预期差 开始Ultramanhu调整IDE Q Boy从头开始看题 我的任务是倒数看题,最后看的题目是J,I,H,G 我看完J觉得J可做
树状数组模块(个人模板)
树状数组模块 ACM个人模板 POJ 2155 题目测试通过 /** * 树状数组模块 * 下标从0开始 */ typedef long DG_Ran; typedef long DG_Num; const DG_Num DG_MAXN = 1005; //2^n DG_Num LowBit(DG_Num n) { return n & (-n); } //获取父节点索引 DG_Num DGFather(DG_Num n)
线性筛法求质数(素数)表 及其原理
/** * 线性筛法求素数表 * 复杂度: O(n) */ const long MAXP = 1000000; long prime[MAXP] = {0},num_prime = 0; int isNotPrime[MAXP] = {1, 1}; void GetPrime_Init()//初始化调用 { for(long i = 2 ; i < MAXP ; i ++) { if(!
HDU HDOJ 3400 Line belt 解题报告
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3400 这题就是一道简单的两重三分 首先设e点为
Southeastern European 2008 Sky Code 解题报告
又是我们的OJ 题目链接: http://www.cn210.com/onlinejudge/problemshow.php?pro_id=92 Description tancu likes space travels but he is a poor software developer and will never be able to buy his own spacecraft. That is why he is preparing to steal the spacecraft of Petru. There is only one problem - Petru has locked the spacecraft with a sophisticated cryptosystem based on the ID numbers of the stars
GCD Determinant 解题报告
http://www.cn210.com/onlinejudge/problemshow.php?pro_id=98 我们的OJ Description </center> Input The next line contains the numbers of S: x1, x2, ..., xn. It is known that each xi is an integer, 0 ≤ xi ≤ 2*109. The input data set is correct and ends with an end of file. </div> Output Sample Input 2 1 2 3 1 3 9 4 1 2 3 6 Sample Output 1 12 4 首先由于行列式交换
PKU POJ 3757 Simple Distributed storage system 解题报告
题目链接:http://acm.pku.edu.cn/JudgeOnline/problem?id=3757 题目大意 第一行输入n,k,f表示
PKU POJ 2976 Dropping tests 解题报告
题目链接: http://acm.pku.edu.cn/JudgeOnline/problem?id=2976 0-1分数规划 最优比例生成树 迭代法 证明:(前几次都是看别人的,这次自己证明) 对于集合s,令l* = max{ a(x) / b(x) } = a(x*) / b(x*).l为
PKU POJ 2728 Desert King 解题报告
题目链接:http://acm.pku.edu.cn/JudgeOnline/problem?id=2728 和3757一样都是01分数规划的
PKU POJ 1141 Brackets Sequence 解题报告
链接: http://acm.pku.edu.cn/JudgeOnline/problem?id=1141 题目意思是输入一些括号,补充括号使之成为没有错误的括号就是只能有括号组在括号组里面,不能出现([)]或者([)]一类的情况 方法是DP
POJ PKU Let's Go to the Movies 解题报告
题目链接:http://acm.pku.edu.cn/JudgeOnline/problem?id=3513 题目大意是输入树状的家庭关系,问