python显示100以内素数，5个一行

尝试找资料但仍然错误
自己的代码如下：
for i in range(3,101):
for j in range(2, i):
if i%j==0:
break
else:
print(i,end=' ')
count+=1
if count==5:
print(i)
count=0
求指点

n = 0
for i in range(2,101):
    count = 0
    for j in range(2,i):
        if i%j == 0:
            break
        else:
            count += 1
    if count > i-3:
        n += 1
        if n % 5 == 0:
            print(i)
        else:
            print(i,end=" ")

报错贴出来！你count也没有定义！

count = 1
for i in range(3, 101):
for j in range(2, i):
if i % j == 0:
break
else:
if count == 5:
print(i, end='\n')
count = 1
else:
print(i, end=' ')
count += 1

shu=[]
for x in range(3,101):
    for i in range(2,x):
        if x%i==0:
            break
    else:
        shu.append(x)
count=0
for item in shu:
    print(item,end=' ')
    count+=1
    if count % 5 == 0:
        print()

ssList = []
for i in range(2, 101):
# 当前数字是否是素数
ssFlag = True
for j in range(2, i):
if i == 2:
ssList.append("2")
break
if i % j == 0 & j != i: # 满足此条件就不是素数
ssFlag = False
if len(ssList) == 5:
print(" ".join(ssList))
ssList.clear()
if ssFlag is True:
ssList.append(str(i))
if len(ssList) > 0:
print(" ".join(ssList))

def do(num):
    li=[]
    for i in range(3,num):
        for k in range(2,i//2+1):#只循环一半，因为任何一个数除了自身，不会被它一半以上的数整除的
            if i%k==0:
                break
        else:
            li.append(i)
            if len(li)==5:
                print(li)
                li.clear()
    if li:
        print(li)



do(100)