Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Background

We are the champions of the Eurocup! Well... almost. Actually, we lost in first round and didn't win any match. But we were beaten by the real champions so, optimistically, we could be the second! Cup competitions are always unfair...

The Problem

Let's consider a cup competition, that consists of N rounds where 2^N teams take part. The winners of each round proceed to the next round. So in round 1 there are 2^N teams, in round 2 there are 2^N-1 teams, and so on. Teams are numbered 0, 1, 2, ..., 2^N -1. The matches in each round are scheduled according to this order. Assume that, when two teams play, the one with the lowest number always wins. In conclusion, in round 1 the matches are: (0 vs. 1), (2 vs. 3), (4 vs.5), ...; in round 2: (0 vs. 2), (4 vs. 6), ... The situation for N=3 is shown in the picture below.

A team, X, is better than all the teams it has beaten, and it is worse than all the teams it has lost with. Moreover, we can apply a sort of transitivity: X is better than all the team that are worse than the teams that are worse than X; and X is worse than the teams that are better than the teams that are better than X. For example, in the case of N=3, team 7 is worse than 6, 4 and 0.

When the competition is finished, all teams are given a ranking from 1 to 2^N (different for each team, 1 for the winner, 2 for the subchampion, etc.). The ranking of team X is R(X). A valid ranking has the following property: if X is better than Y, then R(X) < R(Y). Obviously, there are many possible valid rankings (because there are pairs of teams that we have no information between them). We define the optimistic classification of a team X as the smallest value of R(X) for any valid ranking R. Similarly, the pessimistic classification of X is the biggest value of R(X) for any valid ranking R.

Given the number of rounds, N, and the number of a team, X, you have to compute the optimistic and pessimistic classification of X.

The Input

The first line of the input contains an integer M, indicating the number of test cases. Each test case consists of two integers N and X. N is the total number of rounds in the competition (0<N<31), and X is the number of a team (0<= X <= 2^N -1).

The Output

For each test case, the output should consist of a single line with two integers: the optimistic and the pessimistic classification of the corresponding team.

Sample Input

4
3 1
4 10
2 2
4 15

Sample Output

2 8
3 15
2 3
5 16

---

如果你把它們全部都轉成二進制的話, 你會發現, 比賽由低位的 bit 開始比較
1 的個數+1 為最佳名次
lowbit 的值也就是他撐到哪一個回合時贏了多少隊

#include <stdio.h>

int main() {
    int t, n, m;
    scanf("%d", &t);
    while(t--) {
        scanf("%d %d", &n, &m);
        int tot = 1<<n;
        int tmp = m, cnt = 0;
        while(tmp)
            cnt += tmp&1, tmp >>= 1;
        tmp = (m == 0) ? (1<<n) : (m&-m);
        printf("%d %d\n", cnt+1, tot-tmp+1);
    }
    return 0;
}