001串的熵 - 蓝桥云课 (lanqiao.cn)
import mathn = 23333333for i in range(1, n >> 1):j = n - ia = -(i / n) * (math.log2(i / n)) * i - (j / n) * (math.log2(j / n)) * ja = round(a, 4)if a == 11625907.5798:print(i)break
0求和 - 蓝桥云课 (lanqiao.cn)
n=20230408 print(n*(n+1)>>1)
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