7143. Treasure

Constraints

Time Limit: 1 secs, Memory Limit: 256 MB

Description

One day, MXWai found a treasure map. The map was
divided into 2^n * 2^n grids. The top left corner was (0, 0) and the down right corner was (2^n-1, 2^n-1). The grids and their coordinates are displayed below.

Moreover, she found a secret sequence of length
n, and a secret article saying that

"Repeat
these operations n times then you will find the treasure.

1、divide
the map into four 2^(n-1) * 2^(n-1) parts: the top left part labeled 0, the top right part labeled 1, the down left part labeled 2 and the down right part labeled 3

2、select
the corresponding part according to the first digit of the sequence

3、erase
the first digit of the sequence

4、decrease
n by one

finally
you will find out where the treasure is"

You know MXWai is just a majia, so you know she
is stupid. Now it is your task to find out the location of the treasure.

Input

An integer T (1<=T<=1000000), indicating the number
of test cases.

For each test case, there is only one line, containing
an integer n (1<=n<=31), and a sequence of length n (the sequence only contains digits ‘0’, ‘1’, ‘2’ and ’3’).

Output

Output two integers, indicating the coordinates
of the treasure.

Sample Input

3
1 3
2 00
2 01

Sample Output

1 1
0 0
0 1

Hint

This problem has huge input data, use scanf() instead
of cin to read data to avoid time limit exceed.

Problem Source

“星海通杯”第四届中山大学ICPC新手赛 by 黄文浩

这道题其实就是无限划分。。

#include <string.h>
#include <vector>
#include <queue>
#include <stdio.h>
#include <stack>
#include <math.h>
using namespace std;

int main() {
    int t, n;
    scanf("%d", &t);
    while (t--) {
        char dir[35] = {0};
        long long i_min = 0, j_min = 0, i_max, j_max;
        scanf("%d %s", &n, dir);
        i_max = (long long)pow(2, n) - 1;
        j_max = (long long)pow(2, n) - 1;
        for (int i = 0; i < n; i++) {
            if (dir[i] == '0') {
                i_max = i_min + (i_max - i_min) / 2;
                j_max = j_min + (j_max - j_min) / 2;
            } else if (dir[i] == '1') {
                i_max = i_min + (i_max - i_min) / 2;
                j_min = j_max - (j_max - j_min) / 2;
            } else if (dir[i] == '2') {
                i_min = i_max - (i_max - i_min) / 2;
                j_max = j_min + (j_max - j_min) / 2;
            } else {
                i_min = i_max - (i_max - i_min) / 2;
                j_min = j_max - (j_max - j_min) / 2;
            }
        }
        printf("%lld %lld\n", i_min, j_min);
    }
    return 0;
}