当前位置: 首页 > 面试经验 >

2022/7/23 科大讯飞算法岗笔试

优质
小牛编辑
131浏览
2023-03-28

2022/7/23 科大讯飞算法岗笔试




1.排序然后看相邻的元素之差是否大于4,太水了。。。
n=int(input())
a=[]
for i in range(n):
    a.append(int(input()))
a.sort()
flag=True
for i in range(1,len(a)):
    if a[i]-a[i-1]>4:
        flag=False
        break
a=[str(v) for v in a]
print(' '.join(a))
if flag:
    print(1)
else:
    print(0)

2.leetcode 原题 基本计算器III 
def solve(s):
    global idx
    stk=[]
    num=0
    sign='+'

    f=lambda x:ord(x)-ord('0')
    while idx<len(s):
        c=s[idx]
        if c.isdigit():
            num=num*10+f(c)
        if c=='(':
            idx+=1
            num=solve(s)
        if (not c.isdigit() and c!=' ')&nbs***bsp;idx==len(s)-1:
            pre=0
            if sign=='+':
                stk.append(num)
            elif sign=='-':
                stk.append(-num)
            elif sign=='*':
                pre=stk.pop()
                stk.append(pre*num)
            elif sign=='/':
                pre=stk.pop()
                stk.append(pre//num)
            sign=c
            num=0
        if c==')':
            break
        idx+=1
    return sum(stk)

idx=0
s=input()
print(solve(s))
3.给9个样本,利用朴素贝叶斯对接下来输入的样本进行分类
buf=[[1,2,0,2,1],
     [0,1,1,2,1],
     [1,0,1,1,1],
     [1,2,1,0,1],
     [1,0,0,1,0],
     [0,0,0,1,0],
     [0,1,1,1,0],
     [0,0,0,0,0],
     [1,0,0,0,0]]
px=[[[0]*2 for j in range(3)] for i in range(4)]
for i in range(4):
    for j in range(3):
        cnt1=0
        cnt0=0
        for v in buf:
            if v[i]==j and v[-1]==1:
                cnt1+=1
            if v[i]==j and v[-1]==0:
                cnt0+=1
        px[i][j][1]=cnt1/4
        px[i][j][0]=cnt0/5
x=eval(input())
ans=[4/9,5/9]
for i in range(4):
    for k in range(2):
        ans[k]*=px[i][x[i]][k]
if ans[0]>ans[1]:
    print(0)
else:
    print(1)




#科大讯飞##科大讯飞笔试##投票#
 类似资料: