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此题有O(q+mlogm)的算法（设序列中不同值的个数为m）每次询问不需要O(logn)复杂度。 只要从小到大记录这个序列每个值的起始位置、结束位置和频数（即离散化），对频数进行rmq即可，每次询问只用O(1)的复杂度，300多ms就可以过了。 Followed by:
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