Boring Game

题意

有 n 张纸叠在一起平铺在桌面上，现在把这些纸向右折 k 次。

变成左图的样子，现在给出 \(2 * n*2^k\)个数字，依次从上往下放成左图的样子，让从上到下一行一行输出还原之后的数字序列。

题解

比赛的时候就是没有想到模拟，一步一步的还原。

代码

#include <bits/stdc++.h>
#define fuck system("pause")
#define emplace_back push_back
#define pb push_back
using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;
const double eps = 1e-6;
const int inf = 0x3f3f3f3f;
const int N = 2e6 + 10;

int arr[N], tmp[N];
void solve(int n, int len)
{
    for (int i = 1; i <= n; i++) {
        tmp[i] = arr[i];
    }
    int cnt = 0;
    for (int i = n - len + 1; i > 0;i-=len){
        for (int j = i + len / 2 - 1; j >= i;j--){
            arr[++cnt] = tmp[j];
        }
    }
    for (int i = 1; i <= n; i += len) {
        for (int j = i + len / 2; j < i + len;j++){
            arr[++cnt] = tmp[j];
        }
    }
}
int main()
{
    int T;
    scanf("%d", &T);
    while (T--) {
        int n, k;
        scanf("%d%d", &n, &k);
        int len = 2 * n * 1 << k;
        for (int i = 1; i <= len; i++) {
            scanf("%d", &arr[i]);
        }
        int now = len;
        for (int i = 1; i <= k; i++) {
            solve(len, now);
            now >>= 1;
        }
        int r = len / (1 << k);
        int c = 1 << k;
        int cnt = 0;
        for (int i = 1; i <= r; i++) {
            for (int j = 1; j <= c;j++){
                tmp[++cnt] = arr[i + (j - 1) * r];
            }
        }
        printf("%d", tmp[1]);
        for (int i = 2; i <= len;i++){
            printf(" %d", tmp[i]);
        }
        printf("\n");
    }
    return 0;
}