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 }