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为了解决这个问题，我们需要计算最少需要使用第二个机器的次数才能生成目标项链。这个问题可以通过预处理回文区间并使用贪心算法来解决。

方法思路

解决代码

def manacher(s):
    n = len(s)
    a = ['!$'] + ['$'] * n + ['$']
    a[1] = s[0]
    a[2] = '$'
    for i in range(2, 2 * n):
        a[i] = s[i - 2]
        a[i + 1] = '$'
    a[2 * n] = '*'
    p = [0] * (2 * n + 2)
    p[1] = 1
    rmax = 0
    id = 1
    for i in range(2, 2 * n + 2):
        if i > rmax:
            p[i] = 1
        else:
            k = i - p[i - 1]
            if a[k] == a[i + p[i - 1] - 1]:
                p[i] = p[i - 1] + 2
            else:
                p[i] = 1
        if i + p[i] - 1 > rmax:
            rmax = i + p[i] - 1
            id = i
    d = [(0, 0)] * (2 * n + 2)
    for i in range(1, 2 * n + 1):
        d[i] = (i - p[i], i + p[i] - 1)
    events = []
    for i in range(1, 2 * n + 1):
        if d[i][0] == 0 and d[i][1] == n - 1:
            events.append((d[i][1], 'event'))
        elif d[i][1] < n:
            events.append((d[i][1], 'interval'))
    events.sort()
    res = 0
    last_end = -1
    for event in events:
        if event[0] > last_end:
            res += 1
            last_end = event[0]
    return res
def solve(s):
    n = len(s)
    if n == 0:
        return 0
    events = []
    for i in range(n):
        l = i - manacher(s)[i]
        r = i + manacher(s)[i]
        events.append((r, i, i))
    events.sort()
    res = 0
    last = -1
    for r, l, i in events:
        if r > last:
            res += 1
            last = r
    return res - 1
def main():
    import sys
    for line in sys.stdin:
        s = line.strip()
        print(solve(s))
if __name__ == "__main__":
    main()

代码解释

该方法通过预处理回文区间并使用贪心算法，确保了在最少的连接次数下生成目标项链。

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