I'm WA !wrong answer!

wzk2013L @ 2024-11-10 21:52:58

by King_and_Grey @ 2024-11-10 21:58:13

@wzk2013 不是，题解。。。

import math#引入math库
a = int(input())   #以整数方式来读入
#lee=len(str(pow(2,a))) len()表示字符串长度，str()表示强制转换为字符串，这样求字符串长度会超时（70'）
lee = math.log10(2)*a+1   #求长度
b=pow(2,a,100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000)   #1后面499个0，注意不要使用1e499，会RE
b = b-1   #对输出结果减1
c ='%0500d' %b   #格式化数字补0，转为字符串
a1=int(50)
a2 =int(10)
print(int(lee))
k =int(0)
for j in range(a2):

    for i in range(a1):
        print(c[k],end='')   #对字符串切片输出
        k=k+1
    print(" ")   #换行

by King_and_Grey @ 2024-11-10 21:59:23

@wzk2013 C++要高精度 and 快速幂