题目链接:http://acm.hdu.edu.cn/showproblem.php? pid=1038 Biker's Trip Odometer Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5226    Accepted Submission(s): 3476 Problem Description Most bicycle…

题目真的考验英语 题目链接 需要进行单位的转换 对于Pi用:3.1415927. 5280步相当于1英里. 12英寸相当于1步. 60分钟等于1小时 60秒等于1分钟. 201.168米等于1弗朗.(根本没用这个!吓唬人的!) #include <iostream> #define Pi 3.1415927 using namespace std; int main() { double d,c,t; ; while (cin>>d>>c>>t, c) {…

Spy's Work Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1266    Accepted Submission(s): 388 Problem Description I'm a manager of a large trading company, called ACM, and responsible for the…

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Higher Math Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2219    Accepted Submission(s): 1219 Problem Description You are building a house. You’d prefer if all the walls have a precise right…

这里是杭电hdu上的链接:http://acm.hdu.edu.cn/showproblem.php?pid=3999  Problem Description: As we know,the shape of a binary search tree is greatly related to the order of keys we insert. To be precisely: 1.  insert a key k to a empty tree, then the tree becom…

Problem Description 谁画8画的好,画的快,今后就发的快,学业发达,事业发达,祝大家发,发,发. Input 输入的第一行为一个整数N,表示后面有N组数据. 每组数据中有一个字符和一个整数,字符表示画笔,整数(>=5)表示高度. Output 画横线总是一个字符粗,竖线随着总高度每增长6而增加1个字符宽.当总高度从5增加到6时,其竖线宽度从1增长到2.下圈高度不小于上圈高度,但应尽量接近上圈高度,且下圈的内径呈正方形. 每画一个"8"应空一行,但最前和最后都无空…

Eddy's research I Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5793    Accepted Submission(s): 3459 Problem Description Eddy's interest is very extensive, recently he is interested in prime n…

分析:水题. #include<iostream> using namespace std; #define N 5050 char a[N],b[N],tmp[N]; void Read(char p[]) { getchar(); gets(tmp); while(gets(tmp)) { if(strcmp(tmp,"END")==0) break; if(strlen(tmp)!=0) strcat(p,tmp); strcat(p,"\n");…

>>点击进入原题测试<< 思路:线段树水题,可以手敲 #include<string> #include<iostream> #include<algorithm> #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 using namespace std; ; ]; void build(int l, int r, int rt) { if (l == r){ cin &g…