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将N^4改进为N^3*logN算法.对每行的B值小到大排序, N^2 dp求出f[i][j], 表示i行j~num[i]的最大P值. N^2枚举B值 * N枚举每行 * logN二分查找大于等于当前B的位置,即f[i][x]. 嫌麻烦的话用lower_bound就OK了. Followed by:
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