UVA - 10795 A Different Task

最新推荐文章于 2021-01-20 21:52:48 发布

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最新推荐文章于 2021-01-20 21:52:48 发布

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原文链接：http://www.cnblogs.com/wangsong/p/7546213.html

本文介绍了一种特殊的汉诺塔问题，并提供了一段C++代码来解决该问题。通过对输入的不同盘子数量及初始状态进行分析，程序能够计算出将所有盘子从一个柱子移动到另一个柱子所需的最少步骤。

Input

The input le contains at most 100 test cases. Each test case starts with a positive integer N (1 ≤
N ≤ 60), which means the number of disks. You will be given the arrangements in next two lines. Each
arrangement will be represented by N integers, which are 1, 2 or 3. If the i-th (1 ≤ i ≤ N) integer is
1, you should consider that i-th disk is on Peg-A. Input is terminated by N = 0. This case should not
be processed.
Output
Output of each test case should consist of a line starting with `Case #: ' where # is the test case
number. It should be followed by the minimum number of moves as specied in the problem statement.
Sample Input
3
1 1 1
2 2 2
3
1 2 3
3 2 1
4
1 1 1 1
1 1 1 1
0
Sample Output
Case 1: 7
Case 2: 3
Case 3: 0

#include <cstdio>

using namespace std;

int A[70], B[70];

long long int f(int *A, int i, int x);

int main() {
    int n, t = -1;
    for (int base = 1; scanf("%d", &n) && n; t = -1) {
        for (int i = 0; i < n; ++i) {
            scanf("%d", A + i);
        }
        for (int i = 0; i < n; ++i) {
            scanf("%d", B + i);
            if (A[i] != B[i])
                t = i;
        }
        int x = 6 - A[t] - B[t];
        printf("Case %d: %lld\n", base++, t == -1 ? 0 : (f(A, t - 1, x) + f(B, t - 1, x) + 1));
    }
}

long long f(int *A, int i, int x) {
    if (i < 0)
        return 0;
    if (A[i] == x)
        return f(A, i - 1, x);
    return f(A, i - 1, 6 - x - A[i]) + (1LL << i);  //这里1LL要注意了
}