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Ubuntu下搭建nodejs+express+mongodb环境简单教程 -
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基德KID.1412:
su1216 写道竖线代表或者,不代表替换
对哦~ 谢谢你的提 ...
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su1216:
竖线代表或者,不代表替换
正则表达式中特殊字符的用法(收藏) -
qiqijianglu:
基德KID.1412 写道qiqijianglu 写道基德KI ...
【高斯消元 求期望】HDU 4418 Time travel
文章列表
http://acm.hdu.edu.cn/showproblem.php?pid=1385
题意: 找一个路径使得从一个地方坐的士到另一个地方花费最小,其中除了起点和终点,途中经过的站点要收费
Sample Input
5
0 3 22 -1 4
3 0 5 -1 -1
22 5 0 9 20
-1 -1 9 0 4
4 -1 20 4 0
5 17 8 3 1
1 3
3 5
2 4
-1 -1
0
Sample Output
From 1 to 3 :
Path: 1-->5-->4-->3
Total cost : 21
From 3 to 5 :
Path: ...
http://acm.hdu.edu.cn/showproblem.php?pid=1142
Problem Description
Jimmy experiences a lot of stress at work these days, especially since his accident made working difficult. To relax after a hard day, he likes to walk home. To make things even nicer, his office is on one side of a forest, and his ...
http://acm.hdu.edu.cn/showproblem.php?pid=1575
Problem Description
A为一个方阵,则Tr A表示A的迹(就是主对角线上各项的和),现要求Tr(A^k)%9973。
Input
数据的第一行是一个T,表示有T组数据。
每组数据的第一行有n(2 <= n <= 10)和k(2 <= k < 10^9)两个数据。接下来有n行,每行有n个数据,每个数据的范围是[0,9],表示方阵A的内容。
Output
对应每组数据,输出Tr(A^k)%9973。
Sample Input
2
2 2
1 0
0 1
3 ...
http://acm.hdu.edu.cn/showproblem.php?pid=1217
题意:给出各种钱之间的兑换机制,求不断兑换后是否可以产生利润?
对于第一个案例:start with 1 US dollar and buy 0.5 * 10.0 * 0.21 = 1.05 US dollars, making a profit of 5 percent
Sample Input
3
USDollar
BritishPound
...
http://poj.org/problem?id=1679
题意:问最小生成树是否唯一
思路:
用Kruskal先求最小生成树,结果即为min,把所用到的边记录下来(这里是记录的对应的下表),然后枚举这些边,
每次去掉一个边再求一次最小生成树,结果为tmin,如果能构成最小生成树tmin==min,则说明最小生成树不唯一
Sample Input
2
3 3
1 2 1
2 3 2
3 1 3
4 4
1 2 2
2 3 2
3 4 2
4 1 2
Sample Output
3
Not Unique!
代码自己YY吧……
#include <iostream>
...
http://acm.hdu.edu.cn/showproblem.php?pid=2647
题意:先给出人数n和关系数m,再给出工人之间的工资大小关系,例如给出a b表示a>b,求老板至少要给多少工资【注:工人至少有888元的工资】
Sample Input
2 1
1 2
2 2
1 2
2 1
6 5
2 4
3 6
3 5
1 2
1 3
4 3
1 2
3 4
2 3
Sample Output
1777
-1
5532
3558
#include <iostream>
#include <queue>
using namespace st ...
http://acm.hdu.edu.cn/showproblem.php?pid=1695
题意:求[a,b]和[c,d]中分别取一个数,问取到的两个数的gcd=k的对数!!其中(2,3)跟(3,2)这2种类型只算一种,视为重复
Sample Input
2
1 3 1 5 1
1 11014 1 14409 9
Sample Output
Case 1: 9
Case 2: 736427
#include <iostream>
#include <fstream>
#include <algorithm>
#include <st ...
http://acm.hdu.edu.cn/showproblem.php?pid=1285
Problem Description
有N个比赛队(1<=N<=500),编号依次为1,2,3,。。。。,N进行比赛,比赛结束后,裁判委员会要将所有参赛队伍从前往后依次排名,但现在裁判委员会不能直接获得每个队的比赛成绩,只知道每场比赛的结果,即P1赢P2,用P1,P2表示,排名时P1在P2之前。现在请你编程序确定排名。
Input
输入有若干组,每组中的第一行为二个数N(1<=N<=500),M;其中N表示队伍的个数,M表示接着有M行的输入数据。接下来的M行数据中,每行也 ...
http://acm.hdu.edu.cn/showproblem.php?pid=1496
题意:求有多少个解!!
Problem Description
Consider equations having the following form:
a*x1^2+b*x2^2+c*x3^2+d*x4^2=0
a, b, c, d are integers from the interval [-50,50] and any of them cannot be 0.
It is consider a solution a system ( x1,x2,x3,x4 ) that veri ...
http://acm.hdu.edu.cn/showproblem.php?pid=1348
典型凸包题,求外围城墙的周长
Sample Input
1
9 100
200 400
300 400
300 300
400 300
400 400
500 400
500 200
350 200
200 200
Sample Output
1628
#include <iostream>
#include <fstream>
#include <algorithm>
#include <string>
#include <se ...
http://acm.hdu.edu.cn/showproblem.php?pid=1086
题意:求一堆线段两两相交的次数,即使交点重叠也算在内
更详细的几何讲解:http://dev.gameres.com/Program/Abstract/Geometry.htm#判断两线段是否相交
Sample Input
2
0.00 0.00 1.00 1.00
0.00 1.00 1.00 0.00
3
0.00 0.00 1.00 1.00
0.00 1.00 1.00 0.000
0.00 0.00 1.00 0.00
0
Sample Output
1
3
#include & ...
http://poj.org/problem?id=2318
题意:求每个被割线分离的区间内各有多少个点?
Sample Input
5 6 0 10 60 0
3 1
4 3
6 8
10 10
15 30
1 5
2 1
2 8
5 5
40 10
7 9
4 10 0 10 100 0
20 20
40 40
60 60
80 80
5 10
15 10
25 10
35 10
45 10
55 10
65 10
75 10
85 10
95 10
0
Sample Output
0: 2
1: ...
此题乃师兄[ TT_last ]原创题!Orz膜拜一下
a simple problem
Problem Description
As we know , to caculate the sum ofis hard if the n is large,because today is Valentines Day,to make this problem more easy ,you only need to caculate the sum mod c
Input
There are multiply testcases. Each testcase, there is one line ...
http://poj.org/problem?id=1061
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1657
两题一模一样,只是无解时输出情况不同
首先由题意有【x+ms与y+ns建立等价关系,设次数为s】:
x+ms ≡ (y+ns)(mol L)
->(x+ms)%L = (y+ns)%L
->((x+ms) - (y+ns))%L = 0
->(x+ms) - (y+ns) = k*L
化简得:
k*L + (n-m)*s = x-y
令a = L,b = n-m,n = x-y得
ak + ...
http://poj.org/problem?id=2478
Sample Input
2
3
4
5
0
Sample Output
1
3
5
9
求的是:sum(n) = phi(1) + phi(2) + phi(3) + ... + phi(n);
更多欧拉函数的说明:http://972169909-qq-com.iteye.com/blog/1131309
#include <iostream>
#include <fstream>
#include <algorithm>
#include <string>
# ...