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反向DP的,f[i][j]表示在第i种面值钞票取j张时最接近cash的值,方程:f[i][j] = max(f[i+1][k]+j*d[i]-cash) 约束条件f[i+1][k]+j*d[i]-cash>0, O(n*m^2)的DP, DP之前对n[]的排序,请问方程有问题么?Followed by:
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