2014
01-05

# The Final Countdown

When Doctor Doom modeled LAI-BACH (the Latverian Aerospace Institute, Bolograd Active Control Headquarters) on Mission Command with
NASA, he perhaps cribbed from the source a bit too closely.

NASA (and therefore LAI-BACH) have a peculiar way of handling countdowns to missions. While the clock may state that it is, say, thirty minutes
until liftoff for a rocket or shuttle, there are built-in holds where the clock is stopped for a set amount of time. For example, there may be a 15-minute
hold when the countdown reaches the 8-minute mark. The end result is that more time passes from the beginning of the countdown to the end than
the numbers on the clock would imply.

LAI-BACH uses a simple "command file" format to manage holds for their countdowns on their computer systems. All commands start with a
positive integer no greater than 1440 (countdowns longer than a day make Doctor Doom angry), followed by a directive. The format of the
commands is as follows:

All conditions are represented by short strings of lowercase letters, no more than 20 characters long; their actual values are not provided in the
command structure, as they are determined by actual conditions during the launch (whether the fuel tanks have been topped off, whether the
Fantastic Four is currently fighting Doctor Doom, and so on), but they are always either true or false and do not change during the course of a single
countdown. A hold time n is a positive integer no more than 60 (minutes), as long delays also irritate Victor.

Given a particular command file, you are to determine both the shortest possible time and the longest possible time that the countdown can run. The
commands may be present in any order, but there will always be one and only one START directive; there will be precisely as much whitespace on
each line as dictated by the format given above. No two commands in a command file will reference the same minute t.

Input to this problem will begin with a line containing a single integer N (1 ≤ N ≤ 100) indicating the number of data sets. Each data set consists of
the following components:

A line containing a single integer L (1 ≤ L ≤ 100) indicating the number of lines in the particular "command file"; and
A series of L lines representing the command file, in the format described above.

Input to this problem will begin with a line containing a single integer N (1 ≤ N ≤ 100) indicating the number of data sets. Each data set consists of
the following components:

A line containing a single integer L (1 ≤ L ≤ 100) indicating the number of lines in the particular "command file"; and
A series of L lines representing the command file, in the format described above.

2
3
30 START
15 HOLD 5
10 HOLD 5 IF fantasticfour
3
150 HOLD 30 IF NOT fueled
300 START

35 TO 40
300 TO 350

HOJ 2359/POJ 2393 Yogurt
factory

http://poj.org/problem?id=2393

long不能用%I64d。

#include
<cstdio>
#include
<algorithm>
using
namespace
std;
typedef
long
long
int64;
int
main() {
int n, s;
while (scanf(“%d %d”, &n, &s) == 2) {
int64 ans = 0, t;
int
c, y;
for
(int
i = 0; i < n; i++) {
scanf(“%d %d”, &c, &y);
if
(i == 0) {
t = (int64)c;
} else {
t = min(t + s, (int64)c);
}
ans += t * y;
}
printf(“%lld\n”, ans);
}
return 0;
}
HOJ 2490/POJ 3069 Saruman’s
Army

http://poj.org/problem?id=3069

+ r范围内最有的点，再以该点为中心向右，这样找到的是第一个圆圈所能包含的最多的点，以此类推。

#include
<iostream>
#include
<algorithm>
using
namespace
std;
const
int
MAX = 1024;
int
main() {
int r, n, x[MAX];
while (cin >>
r >> n
&&
!(r == -1 && n
== -1)) {
for
(int
i = 0; i < n; i++) {
cin >>
x[i];
}
sort(x, x + n);
int
ans = 0, i = 0;
while
(i < n) {
int s
= x[i];
while
(i < n &&
x[i] <= s + r) {
i++;
}
ans++;
s = x[i - 1];
while
(i < n &&
x[i] <= s + r) {
i++;
}
}
cout <<
ans <<
endl;
}
return 0;
}

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2. 站长，你好！
你创办的的网站非常好，为我们学习算法练习编程提供了一个很好的平台，我想给你提个小建议，就是要能把每道题目的难度标出来就好了，这样我们学习起来会有一个循序渐进的过程！

3. 其实国内大部分公司对算法都不够重视。特别是中小型公司老板根本都不懂技术，也不懂什么是算法，从而也不要求程序员懂什么算法，做程序从来不考虑性能问题，只要页面能显示出来就是好程序，这是国内的现状，很无奈。