答案可能大于 ``long long`` 可以存储的范围。
by dead_X @ 2021-10-10 13:55:07
@[dead_X](/user/111055) 所以咋办!(抱住大腿
by 張懌宸Eason @ 2021-10-10 13:55:53
~~看题解~~
by 褚橙 @ 2021-10-10 13:57:07
@[張懌宸Eason](/user/487283) 可以用数组来存数,然后模拟
by 褚橙 @ 2021-10-10 14:00:03
> 没学过高精
[那就去学](https://oi-wiki.org/math/bignum/)
by cmaths @ 2021-10-10 14:00:18
Learn how to use Python3.
by Eason_AC @ 2021-10-10 14:00:53
```python3
n=int(input())
print(pow(2,n)-1)
```
by dead_X @ 2021-10-10 14:01:03
必须得用高精度算法呀(位运算也可以),因为n<=15000
给你一个高精度乘法模板吧
string s1,s2;
cin>>s1>>s2;
int len1=s1.size();
int len2=s2.size();
for(int i=0;i<len1;i++) a[len1-i-1]=s1[i]-'0';
for(int i=0;i<len2;i++) b[len2-i-1]=s2[i]-'0';
for(int i=0;i<len1;i++)
for(int j=0;j<len2;j++){
c[i+j]+=a[i]*b[j];
c[i+j+1]+=c[i+j]/10;
c[i+j]%=10;
}
int p=len1+len2;
while(c[p]==0&&p>0) p--;
for(int i=p;i>=0;i--) cout<<c[i];
位运算就很简单,不过有点不适合新手,就不多说了啊
by zhubicheng @ 2021-10-10 14:01:28
@[Eason_AC](/user/112917) 出现了大大大大大大佬!
(我其实会一点PythonDOGE)
by 張懌宸Eason @ 2021-10-10 14:07:41
@[zhubicheng](/user/360410) 谢谢!!!
by 張懌宸Eason @ 2021-10-10 14:08:02