听说是邀请赛啊,大概做了做…中午出去吃了个饭回来过掉的I。然后去做作业了……

 #include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath> using namespace std; #define fr first
#define sc second
#define pb(a) push_back(a)
#define Rint(a) scanf("%d", &a)
#define Rll(a) scanf("%I64d", &a)
#define Rs(a) scanf("%s", a)
#define FRead() freopen("in", "r", stdin)
#define FWrite() freopen("out", "w", stdout)
#define Rep(i, len) for(int i = 0; i < (len); i++)
#define For(i, a, len) for(int i = (a); i < (len); i++)
#define Cls(a) memset((a), 0, sizeof(a))
#define Full(a) memset((a), 0x7f7f, sizeof(a)) typedef long long ll;
const int maxn = ;
string s;
int g[]; int main() {
// FRead();
int T;
Rint(T);
while(T--) {
string pre;
int pos = ;
Cls(g);
while(cin >> s) {
if(s[] == 'E') break;
if(::s == pre) {
pos = - pos;
g[pos]++;
pre = ::s;
}
else {
g[pos]++;
pre = ::s;
}
}
// cout << g[0] << " "<< g[1] <<endl;
printf("%d\n", g[] * g[]);
}
return ;
}

C

 #include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath> using namespace std; #define fr first
#define sc second
#define pb(a) push_back(a)
#define Rint(a) scanf("%d", &a)
#define Rll(a) scanf("%I64d", &a)
#define Rs(a) scanf("%s", a)
#define FRead() freopen("in", "r", stdin)
#define FWrite() freopen("out", "w", stdout)
#define Rep(i, len) for(int i = 0; i < (len); i++)
#define For(i, a, len) for(int i = (a); i < (len); i++)
#define Cls(a) memset((a), 0, sizeof(a))
#define Full(a) memset((a), 0x7f7f, sizeof(a)) typedef long long ll;
ll n, k; int main() {
// FRead();
while(~Rint(n) && ~Rint(k)) {
ll t;
while() {
if(t == n) break;
t = n;
n /= ;
n += k;
}
cout << t << endl;
}
return ;
}

H

 #include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath> using namespace std; #define fr first
#define sc second
#define pb(a) push_back(a)
#define Rint(a) scanf("%d", &a)
#define Rll(a) scanf("%I64d", &a)
#define Rs(a) scanf("%s", a)
#define FRead() freopen("in", "r", stdin)
#define FWrite() freopen("out", "w", stdout)
#define Rep(i, len) for(int i = 0; i < (len); i++)
#define For(i, a, len) for(int i = (a); i < (len); i++)
#define Cls(a) memset((a), 0, sizeof(a))
#define Full(a) memset((a), 0x7f7f, sizeof(a)) const int maxn = ;
double n, V;
double a[maxn], v[maxn];
double ans; int main() {
// FRead();
while(cin >> n >> V) {
bool exflag = ;
ans = ;
Rep(i, n) {
cin >> a[i] >> v[i];
if(abs(v[i]) >= V && a[i] != ) exflag = ;
}
if(exflag) {
printf("Bad Dog\n");
continue;
}
double pos = ;
Rep(i, n) {
a[i] += ans * v[i];
if(a[i] == pos) continue;
if(a[i] < pos) V = - * abs(V);
else V = abs(V);
if(abs(V - v[i]) <= ) {
exflag = ;
break;
}
double d = abs(pos - a[i]);
double t = d / abs(V - v[i]);
ans += t;
pos += t * V;
}
if(exflag) printf("Bad Dog\n");
else printf("%.2lf\n", ans);
}
return ;
}

I

 #include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath> using namespace std; #define fr first
#define sc second
#define pb(a) push_back(a)
#define Rint(a) scanf("%d", &a)
#define Rll(a) scanf("%I64d", &a)
#define Rs(a) scanf("%s", a)
#define FRead() freopen("in", "r", stdin)
#define FWrite() freopen("out", "w", stdout)
#define Rep(i, len) for(int i = 0; i < (len); i++)
#define For(i, a, len) for(int i = (a); i < (len); i++)
#define Cls(a) memset((a), 0, sizeof(a))
#define Full(a) memset((a), 0x7f7f, sizeof(a)) const int maxn = ;
int h[maxn];
int ans;
int n; int check(int d) {
int ret = ;
int tmp = ;
for(int i = ; i <= ; i++) {
tmp = ;
if(h[i] == ) continue;
for(int j = i; j <= ; j+=d) {
if(h[j]) tmp++;
else break;
}
ret = max(tmp, ret);
}
return ret;
} int main() {
// FRead();
int tmp;
while(~Rint(n)) {
Cls(h);
ans = ;
Rep(i, n) {
Rint(tmp);
h[tmp]++;
}
//d = 0
for(int i = ; i <= ; i++) {
ans = max(ans, h[i]);
}
for(int d = ; d <= ; d++) {
ans = max(ans, check(d));
}
cout << ans << endl;
}
return ;
}

J

[HZAU]华中农业大学第四届程序设计大赛网络同步赛的更多相关文章

  1. (hzau)华中农业大学第四届程序设计大赛网络同步赛 G: Array C

    题目链接:http://acm.hzau.edu.cn/problem.php?id=18 题意是给你两个长度为n的数组,a数组相当于1到n的物品的数量,b数组相当于物品价值,而真正的价值表示是b[i ...

  2. 华中农业大学第四届程序设计大赛网络同步赛 G.Array C 线段树或者优先队列

    Problem G: Array C Time Limit: 1 Sec  Memory Limit: 128 MB Description Giving two integers  and  and ...

  3. 华中农业大学第四届程序设计大赛网络同步赛 J

    Problem J: Arithmetic Sequence Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 1766  Solved: 299[Subm ...

  4. 华中农业大学第四届程序设计大赛网络同步赛 I

    Problem I: Catching Dogs Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 1130  Solved: 292[Submit][St ...

  5. 华中农业大学第四届程序设计大赛网络同步赛-1020: Arithmetic Sequence,题挺好的,考思路;

    1020: Arithmetic Sequence Time Limit: 1 Sec  Memory Limit: 128 MB Submit:  ->打开链接<- Descriptio ...

  6. 华中农业大学第五届程序设计大赛网络同步赛-L

    L.Happiness Chicken brother is very happy today, because he attained N pieces of biscuits whose tast ...

  7. 华中农业大学第五届程序设计大赛网络同步赛-K

    K.Deadline There are N bugs to be repaired and some engineers whose abilities are roughly equal. And ...

  8. 华中农业大学第五届程序设计大赛网络同步赛-G

    G. Sequence Number In Linear algebra, we have learned the definition of inversion number: Assuming A ...

  9. 华中农业大学第五届程序设计大赛网络同步赛-D

    Problem D: GCD Time Limit: 1 Sec  Memory Limit: 1280 MBSubmit: 179  Solved: 25[Submit][Status][Web B ...

随机推荐

  1. php正则表达式判断是否为ip格式

    <?php $a = '127.0.0.111'; $b = preg_match("/^\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}$/",$a); ...

  2. Jquery-------获取网页参数

    看如下代码: function getURLParameter(name) { return decodeURI( (RegExp(name + '=' + '(.+?)(&|$)').exe ...

  3. C++对MS SQL Server的操作

    今天因为在做一份C++的期末作业,突然想用C++来链接数据库,实现数据的重复利用,所以就作死去百度搜了一下. 更巧的事情是,一搜居然还有很多搜索结果,然后就照着做了. 做的过程很艰辛,就不一一诉说了, ...

  4. For和While在C和MATLAB中的区别——MATLAB的大坑

    For和while是常见的循环关键字,在许多语言中都是通用的.但是想必不是所有人,都被其中的区别困扰过,尤其是MATLAB“程序员”. x=[,,,,,,]; i=; while i<=leng ...

  5. Ext学习-基础组件介绍

    1.目标    学习对象获取,组件基础,事件模型以及学习ExtJS中的基础组件的应用. 2.内容   1.对象获取   2.组件原理以及基础   3.事件模型   4.常用组件的介绍 3.学习步骤 1 ...

  6. SVN提交错误:working copy is not up-to-date解决方法

    我在项目中删了2个jar,然后SVN提交,一直提交不成功 svn在提交时报错如下图: working copy is not up-to-date svn:commit failed(details ...

  7. Codeforces Round #363 (Div. 2)->C. Vacations

    C. Vacations time limit per test 1 second memory limit per test 256 megabytes input standard input o ...

  8. 正确使用stl vecotr erase函数

    erase函数要么删作指定位置loc的元素,要么删除区间[start, end)的所有元素. 返回值是指向删除的最后一个元素的下一位置的迭代器 Parameters All parameters ar ...

  9. 【系统Configmachine.config与自己的应用程序的App.config/Web.Config配置节点重复】解决方法

    自己的应用程序的App.config或Web.Config文件中与系统的C:\Windows\Microsoft.NET\Framework64\v4.0.30319\Configmachine.co ...

  10. Feature Flag

    know more from here: https://www.youtube.com/watch?v=WMRjj06R6jg&list=UUkQX1tChV7Z7l1LFF4L9j_g F ...