1000000000以内的满足条件、误差小于1e-07的(a, b)组合:

1: a=71173049, b=82256448, a*pi-b*e=520.0000000000, |error|=0.0000000000

2: a=190751890, b=220456985, a*pi-b*e=520.0000000000, |error|=0.0000000000

3: a=192575648, b=222564752, a*pi-b*e=520.0000000000, |error|=0.0000000000

4: a=194399406, b=224672519, a*pi-b*e=520.0000000000, |error|=0.0000000000

5: a=196223164, b=226780286, a*pi-b*e=520.0000000000, |error|=0.0000000000

6: a=198046922, b=228888053, a*pi-b*e=520.0000000000, |error|=0.0000000000

7: a=310330731, b=358657522, a*pi-b*e=520.0000000000, |error|=0.0000000000

8: a=315802005, b=364980823, a*pi-b*e=520.0000000000, |error|=0.0000000000

9: a=317625763, b=367088590, a*pi-b*e=520.0000000000, |error|=0.0000000000

10: a=323097037, b=373411891, a*pi-b*e=520.0000000000, |error|=0.0000000000

从数学上对代码里的遍历做了优化：

import math

tolerance = 1E-7

i, N = 0, 1_000_000_000

pice, e520 = math.pi / math.e, 520 / math.e

results = []

for a in range(1, N):

    apice = a * pice - e520

    for b in range(int(round(apice - 1)), int(round(apice + 1)), 1):

        value = a * math.pi - b * math.e

        error = abs(value - 520)

        if error <= tolerance:

            i += 1

            print(f'{i}:\ta={a:<10d}, b={b:<10d}, a*pi-b*e={value:.010f}, er
r={error:.010f}')

            results.append((a, b, error))

# 按误差大小降序输出

results = sorted(results, key=lambda x: x[2])

print(f"{N}以内的满足条件、误差小于{tolerance}的(a, b)组合:")

for i, (a, b, err) in enumerate(results):

    print(f'{i+1:>2d}: a={a:<8d}, b={b:<8d}, a*pi-b*e={a*math.pi-b*math.e:.0
10f}, |error|={err:.010f}')



【 在 iMx 的大作中提到: 】
: a=689, b=605, err=-0.0031678943904142822597998255775819932582954300490
: a=1344, b=1362, err=0.0006760633930817191139759780519944224576264086960
: a=6905, b=7789, err=0.0001111696623166836279987936447473198838605590950      
: ...................
--
FROM 119.233.241.*