摘要: 题目意思很难得懂,给N个2次函数,求最大值中的最小值3分法 (用黄金分割比例可以省去一次计算使速度更快)用CIN COUT会超时Error CurvesTime Limit : 4000/2000ms (Java/Other)Memory Limit : 65536/65536K (Java/Other)Total Submission(s) : Accepted Submission(s) :Problem DescriptionJosephina is a clever girl and addicted to Machine Learning recently. Shepays muc. 阅读全文
posted @ 2013-04-24 21:27 码代码的猿猿 阅读(228) 评论(0) 推荐(0) 编辑
摘要: 男人8题:第8题二分+贪心判断答案Elevator Stopping PlanTime Limit:1000MSMemory Limit:30000KTotal Submissions:1839Accepted:614DescriptionZSoft Corp. is a software company in GaoKe Hall. And the workers in the hall are very hard-working. But the elevator in that hall always drives them crazy. Why? Because there is on 阅读全文
posted @ 2013-04-24 15:07 码代码的猿猿 阅读(372) 评论(0) 推荐(0) 编辑
摘要: #include #include using namespace std;int is_run(int year){ if(year%100!=0&&year%4==0) return 1; if(year%400==0) return 1; else return 0;}int main(){ int k; cin>>k;... 阅读全文
posted @ 2013-04-24 05:31 码代码的猿猿 阅读(186) 评论(0) 推荐(0) 编辑
摘要: 用字符形式输出数字必WA。#include #include using namespace std;int main(){ int n; string s; cin>>n;for(int i=0;i>s; char c=s[0]; int len=s.length(); for(int j=1;j<=len;j++) { if(s[j]==... 阅读全文
posted @ 2013-04-24 03:59 码代码的猿猿 阅读(106) 评论(0) 推荐(0) 编辑
摘要: 充满各种PE的垃圾题#include using namespace std;int n,m;static int c=0;int round;int main(){cin>>round;for(int l=0;l>n>>m){ if(n==0&&m==0) break; int tot=0; int ans; for(int i=1;i<n;i++) fo... 阅读全文
posted @ 2013-04-24 03:30 码代码的猿猿 阅读(156) 评论(0) 推荐(0) 编辑