今年北部某省虽然参加了联考，但存在大面积估分与最终分数差异巨大的情况，网上流传了好几种不同的系数组合，但总有人对不上。
刚巧我拿到了19组各模块正确题数和成绩的对应数据，五大模块对应五元一次方程组，也就是个五阶矩阵嘛，用4*4=16组数据加上满分可以算四次出来了。
然而结果很有意思，这些数据算出来的四组数据都有模块系数为负数的情况，在样本超过10的情况下这样是很不正常的。
所以系数分配一说纯属扯淡，可以确定地说这采取的动态赋分机制，动态赋分顾名思义就是每道题分数不固定，每题根据全省的正确率来算得分系数，简单大家都做对的分低，难度高大家都做错的就分数高。所以常识和数量不到位的朋友们吃了死亏，言语和资料分析这种常规大头项目实际上的性价比贼低。
代码如下：
import numpy as np
# 系数
changshi = 0
yanyu = 0
panduan = 0
shuliang = 0
ziliao = 0
grade = [20,40,35,15,15]
grade1 = [5,32,21,7,12]
grade2 = [7,33,25,11,11]
grade3 = [4,28,31,7,9]
grade4 = [5,26,29,6,10]
grade5 = [8,30,28,7,11]
grade6 = [8,32,29,8,14]
grade7 = [8,31,32,8,11]
grade8 = [5,29,28,6,12]
grade9 = [7,26,28,6,12]
grade10 = [11,33,30,8,10]
grade11 = [9,34,27,5,12]
grade12 = [8,33,26,9,13]
grade13 = [10,31,29,5,13]
grade14 = [8,31,29,4,12]
grade15 = [11,30,27,6,12]
grade16 = (6,31,26,5,10)
m1 = np.array([grade,grade1,grade2,grade3,grade4])
n1 = np.array([100, 53.9, 62.4, 54.8, 52.9])
m2 = np.array([grade,grade5,grade6,grade7,grade8])
n2 = np.array([100, 53.5, 61.4, 62.2, 50])
m3 = np.array([grade,grade9,grade10,grade11,grade12])
n3 = np.array([100, 56, 53.5, 60.4, 62.4])
m4 = np.array([grade,grade13,grade14,grade15,grade16])
n4 = np.array([100, 64.3, 57, 61.1, 55.5])
solution1 = np.linalg.solve(m1, n1)
solution2 = np.linalg.solve(m2, n2)
solution3 = np.linalg.solve(m3, n3)
solution4 = np.linalg.solve(m4, n4)
print("第一组数据")
print(m1)
print(n1)
print(solution1)
print("第一组数据对应系数：常识:%.4f,言语:%.4f,判断:%.4f,数量:%.4f,资料:%.4f" % (solution1[0],solution1[1],solution1[2],solution1[3],solution1[4]))
print("第二组数据")
print(m2)
print(n2)
print(solution2)
print("第二组数据对应系数：常识:%.4f,言语:%.4f,判断:%.4f,数量:%.4f,资料:%.4f" % (solution2[0],solution2[1],solution2[2],solution2[3],solution2[4]))
print("第三组数据")
print(m3)
print(n3)
print(solution3)
print("第三组数据对应系数：常识:%.4f,言语:%.4f,判断:%.4f,数量:%.4f,资料:%.4f" % (solution3[0],solution3[1],solution3[2],solution3[3],solution3[4]))
print("第四组数据")
print(m4)
print(n4)
print(solution4)
print("第四组数据对应系数：常识:%.4f,言语:%.4f,判断:%.4f,数量:%.4f,资料:%.4f" % (solution4[0],solution4[1],solution4[2],solution4[3],solution4[4]))