Codeforces CF#628 Education 8 A. Tennis Tournament
A tennis tournament with n participants is running. The participants are playing by an olympic system, so the winners move on and the losers drop out.
The tournament takes place in the following way (below, m is the number of the participants of the current round):
- let k be the maximal power of the number 2 such that k ≤ m,
- k participants compete in the current round and a half of them passes to the next round, the other m - k participants pass to the next round directly,
- when only one participant remains, the tournament finishes.
Each match requires b bottles of water for each participant and one bottle for the judge. Besides p towels are given to each participant for the whole tournament.
Find the number of bottles and towels needed for the tournament.
Note that it's a tennis tournament so in each match two participants compete (one of them will win and the other will lose).
The only line contains three integers n, b, p (1 ≤ n, b, p ≤ 500) — the number of participants and the parameters described in the problem statement.
Print two integers x and y — the number of bottles and towels need for the tournament.
5 2 3
20 15
8 2 4
35 32
In the first example will be three rounds:
- in the first round will be two matches and for each match 5 bottles of water are needed (two for each of the participants and one for the judge),
- in the second round will be only one match, so we need another 5 bottles of water,
- in the third round will also be only one match, so we need another 5 bottles of water.
So in total we need 20 bottles of water.
In the second example no participant will move on to some round directly.
题意: 有n个人两两对决,每场对决有一个裁判。 每个选手每次参加比赛需要b瓶水,并且裁判需要一瓶。并且选手各需要毛巾p条。 (题意没有明说,需要瞎猜)每个人的毛巾可以循环使用,每个人从比赛到结束都只需要p条毛巾。 问总共需要的水和毛巾数量。
题解: 直接模拟每轮比赛即可。
1 #include <iostream> 2 #include <algorithm> 3 #include <cstdio> 4 using namespace std; 5 6 int n, b, p; 7 8 int main() { 9 scanf("%d%d%d", &n, &b, &p); 10 int bottles = 0, towels = n * p; 11 while(n > 1) { 12 int k = 2; 13 while(k * 2 <= n) k <<= 1; 14 int m = n - k + k / 2; 15 int matches = k / 2; 16 bottles += matches * (2 * b + 1); 17 n = m; 18 } 19 printf("%d %d\n", bottles, towels); 20 return 0; 21 }