modi test

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## itertools

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``` python

# 使用交错级数计算pi

import itertools

from functools import reduce

def pi(N):

    cc = itertools.cycle([4, -4])

    odds = itertools.count(1,2)

    ns = list(itertools.takewhile(lambda x: x <= 2*N-1, odds))

    pp = reduce(lambda x, y: x + y, map(lambda x: next(cc)/x, ns))

    return pp

pi(1000)

pi(100)

```

%% Output

    3.140592653839794

3.1315929035585537

%% Cell type:code id: tags:

``` python

import itertools

def pi(N):

    odd = itertools.count(1, 2)

    o = itertools.takewhile(lambda x: x < 2 * N, odd)

    c = itertools.cycle([4, -4])

    p = sum([next(c) / i for i in o])

    return p

pi(10)

```

%% Output

    3.0418396189294032

3.0418396189294032

%% Cell type:code id: tags:

``` python

100//25

```

%% Output

    4

4

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``` python

a=5

print(a+5)

```

%% Output

10

%% Cell type:code id: tags:

``` python

```