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http://acm.hdu.edu.cn/showproblem.php?pid=5200 Trees Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1285    Accepted Submission(s): 400

Scales Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 512    Accepted Submission(s): 230 Problem Description Give you a scale, a g

B. String time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output One day in the IT lesson Anna and Maria learned about the lexicographic order.St

Backward Digit Sums Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 6223 Accepted: 3602 DescriptionFJ and his cows enjoy playing a mental game. They write down the numbers from 1

本题目是蚂蚁的相遇问题，蚂蚁相遇以后就反向，输出蚂蚁t秒后的次序。 这里用ne数组存下新状态的一切信息。注意交换的次序即可。 因为所有蚂蚁的相对位置不变，所以当前的相对位置也不变。 num存着所有的新位置的相对次序，即当前ne[i].loc是第几大的数字呢？num里面将这个次序分别存下来 order[i]里面则表示 大小顺序序号为i的数据将在第order[i]个输出 之后ne数组又依照数

B. Tower of Hanoi time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output The Tower of Hanoi is a well-known mathematical puzzle. It consists of th

模拟。 注意输入为很多种情况，TD,TH,DH#include<cstdio> #include<cmath> #include<algorithm> #include<cstring> using namespace std;char s1[100],s2[100]; double t,d,h; double a1,a2; double E=2.718281828; int vis[300];v

模拟#include<cstdio> #include<cmath> #include<algorithm> #include<cstring> using namespace std; char ss[10000];int main() { while(gets(ss)) { if(ss[0]=='#') break; int

利用海伦公式由海伦公式得出面积 s=sqrt(p*(p-a)(p-b)(p-c)) p=(a+b+c)/2;由三角形面积公式S=1/2*a*b*sin(C) 和正弦定理a/sin(A)=b/sin(B)=c/sin(C)=d=2*r 得d=a*b*c/2/S; (a,b,c为三边长)周长为d*PI#include<cstdio> #include<cmath> #include<al

模拟。 注意模拟的方法，记录235分别上一次乘的数字。直接暴力会超时。#include<cstdio> #include<cmath> #include<algorithm> #include<cstring> using namespace std; int a[2000]; int n;bool ok(int x) { int flag=1; while(flag) {

FatMouse's Tour Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu Submit Status Practice ZOJ 1105 Appoint description:  System Crawler  (2016-04-21) Descripti

HTML Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu Submit Status Practice ZOJ 1099 Description If you ever tried to read a html document on a Macinto

Octal Fractions Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu Submit Status Practice ZOJ 1086 Description Fractions in octal (base 8) notation can be

C. Pearls in a Row time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output There are n pearls in a row. Let's enume

4 Values whose Sum is 0 Time Limit: 15000MS   Memory Limit: 228000K Total Submissions: 18686   Accepted: 5546 Case Time Limit: 5000MS Description The SUM problem can

Ivan and Powers of Two Time Limit:500MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit Status Description Ivan has got an array of n non-negative integers a1, a2, ...

Encoding Time Limit: 1000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u Submit Status Description Chip and Dale have devised an encryption method to hide their (writ

Digital Roots Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu Submit Status Practice ZOJ 1115 Appoint description:  System Crawler  (2016-04-26) Description

ZOJ_1122: Clock Time Limit: 2000 MS  Memory Limit: 64 MB   64bit IO Format: %lld Submitted: 1  Accepted: 1 [Submit][Status][Web Board] Description You are given a standard 12-hour clock wi

ZOJ_1003: Crashing Balloon Time Limit: 2000 MS  Memory Limit: 64 MB   64bit IO Format: %lld [Submit][Status][Web Board] Description On every June 1st, the Children's Day, there will be a

ZOJ_1009: Enigma Time Limit: 10000 MS  Memory Limit: 32 MB   64bit IO Format: %lld [Submit][Status][Web Board] Description In World War II, Germany once used an electronic encryption mach

ZOJ_1016: Parencodings Time Limit: 2000 MS  Memory Limit: 64 MB   64bit IO Format: %lld Submitted: 0  Accepted: 0 [Submit][Status][Web Board] Description Let S = s1 s2 ... s2n be a well-fo

题目有坑，如果出现了两次借同一本书的操作，那么以后借的一次为开始时间。代码：#include&lt;cstdio&gt; #include&lt;cstring&gt; #include&lt;cmath&gt; int sta[1007]; int n; char s[10]; int time[1007]; using namespace std; int coun,ave; int ca...