【水】TJU Problem 1805 Electrical Outlets

没看懂题,直接看了INPUT,原来真的可以猜出来。

原题:

1805.   Electrical Outlets


Time Limit: 1.0 Seconds   Memory Limit: 65536K
Total Runs: 3321   Accepted Runs: 2575



Roy has just moved into
a new apartment. Well, actually the apartment itself is not very new, even
dating back to the days before people had electricity in their houses. Because
of this, Roy‘s apartment has only one single wall outlet, so Roy can only power
one of his electrical appliances at a time.

Roy likes to watch TV as he works on his computer, and to listen to his HiFi
system (on high volume) while he vacuums, so using just the single outlet is not
an option. Actually, he wants to have all his appliances connected to a powered
outlet, all the time. The answer, of course, is power strips, and Roy has some
old ones that he used in his old apartment. However, that apartment had many
more wall outlets, so he is not sure whether his power strips will provide him
with enough outlets now.

Your task is to help Roy compute how many appliances he can provide with
electricity, given a set of power strips. Note that without any power strips,
Roy can power one single appliance through the wall outlet. Also, remember that
a power strip has to be powered itself to be of any use.

Input specifications

Input vill start with a single integer 1 ≤ N ≤ 20, indicating the number of
test cases to follow. Then follow N lines, each describing a test case. Each
test case starts with an integer 1 ≤ K ≤ 10, indicating the number of power
strips in the test case. Then follow, on the same line, K integers separated by
single spaces, O1 O2 ... OK, where 2 ≤
Oi ≤ 10, indicating the number of outlets in each power strip.

Output specifications

Output one line per test case, with the maximum number of appliances that can
be powered.

Sample Input

3
3 2 3 4
10 4 4 4 4 4 4 4 4 4 4
4 10 10 10 10

Sample Output

7
31
37

Source: Nordic
Collegiate Contest 2005

源代码:

 1 #include <iostream>
 2 using namespace std;
 3
 4 int main()    {
 5     int N;    cin >> N;
 6     while (N--)    {
 7         int n, sum = 0, num;    cin >> n;
 8         for (int i = 0; i < n; i++)    {
 9             cin >> num;    sum += num;
10         }
11         cout << sum - n + 1 << endl;
12     }
13     return 0;
14 } 
时间: 2024-10-13 06:36:16

【水】TJU Problem 1805 Electrical Outlets的相关文章

HDU 2304 Electrical Outlets

Problem Description Roy has just moved into a new apartment. Well, actually the apartment itself is not very new, even dating back to the days before people had electricity in their houses. Because of this, Roy's apartment has only one single wall ou

HDOJ(HDU) 2304 Electrical Outlets(求和、、)

Problem Description Roy has just moved into a new apartment. Well, actually the apartment itself is not very new, even dating back to the days before people had electricity in their houses. Because of this, Roy's apartment has only one single wall ou

TJU Problem 1065 Factorial

注意数据范围,十位数以上就可以考虑long long 了,断点调试也十分重要. 原题: 1065.   Factorial Time Limit: 1.0 Seconds   Memory Limit: 65536KTotal Runs: 6067   Accepted Runs: 2679 The most important part of a GSM network is so called Base Transceiver Station (BTS). These transceiver

TJU Problem 2520 Quicksum

注意: for (int i = 1; i <= aaa.length(); i++) 其中是“ i <= ",注意等号. 原题: 2520.   Quicksum Time Limit: 0.5 Seconds   Memory Limit: 65536KTotal Runs: 2964   Accepted Runs: 1970 A checksum is an algorithm that scans a packet of data and returns a single

TJU Problem 1015 Gridland

最重要的是找规律. 下面是引用 http://blog.sina.com.cn/s/blog_4dc813b20100snyv.html 的讲解: 1 做这题时,千万不要被那个图给吓着了,其实这题就是道简单的数学题. 2 3 首先看当m或n中有一个为2的情况,显然,只需要算周长就OK了.即(m+n-2)*2,考虑到至少其中一个为2,所以答案为2 *m或2*n,亦即m*n.注意这里保证了其中一个数位偶数. 4 当m,n≥3时,考虑至少其中一个为偶数的情况,显然,这种情况很简单,可以得出,结果为m*

TJU Problem 2857 Digit Sorting

原题: 2857.   Digit Sorting Time Limit: 1.0 Seconds   Memory Limit: 65536KTotal Runs: 3234   Accepted Runs: 1704 Several players play a game. Each player chooses a certain number, writes it down (in decimal notation, without leading zeroes) and sorts t

TJU Problem 2101 Bullseye

注意代码中: result1 << " to " << result2 << ", PLAYER 1 WINS."<< endl; 和 result1 << " to " << result2 << ", PLAYER 1 WINS. "<< endl; 虽然只在WINS后只差一个空格,但会导致PE. 原题: 2101.   Bul

TJU Problem 1100 Pi

注: 1. 对于double计算,一定要小心,必要时把与double计算相关的所有都变成double型. 2. for (int i = 0; i < N; i++)         //N 不可写为N - 1,否则当N为1时无法进行: 原题: 1100.   Pi Time Limit: 1.0 Seconds   Memory Limit: 65536KTotal Runs: 5683   Accepted Runs: 2317 Professor Robert A. J. Matthews

TJU Problem 2548 Celebrity jeopardy

下次不要被长题目吓到,其实不一定难. 先看输入输出,再揣测题意. 原文: 2548.   Celebrity jeopardy Time Limit: 1.0 Seconds   Memory Limit: 65536KTotal Runs: 1306   Accepted Runs: 898 It's hard to construct a problem that's so easy that everyone will get it, yet still difficult enough