@[louzezhong20130517](/user/989399) 你这样用通项公式求斐波那契数，估计连longlong都存不下。暴力求解可以实现边算边模，不会超时。
by CSP_juruo @ 2023-08-13 10:43:17
咋算
by louzezhong20130517 @ 2023-08-13 10:44:21
还有他让你模 $2^{31}$，你好像没模
by CSP_juruo @ 2023-08-13 10:44:25
@[cut_subject_person](/user/500031)
by louzezhong20130517 @ 2023-08-13 10:45:17
@[louzezhong20130517](/user/989399) mod
by Argvchs @ 2023-08-13 10:46:17
我试过,可不知道为什么报错了。
by louzezhong20130517 @ 2023-08-13 10:47:18
@[louzezhong20130517](/user/989399)
```cpp
#include<bits/stdc++.h>
#define MOD 2147483648
using namespace std;
long long fb(int n)
{
long long a=1,b=1,c=1;
for (int i=3;i<=n;i++){
c=(a+b)%MOD;
a=b,b=c;
}
return c;
}
void f(int n)
{
int i=2;
cout<<n<<"=";
do{
while(n%i==0)
{
cout<<i;
n/=i;
if(n!=1)
{
cout<<"*";
}
}
i++;
}
while(n!=1);
}
int main(){
long long n;
cin>>n;
f(fb(n));
return 0;
}
```
by CSP_juruo @ 2023-08-13 10:49:42
暴力求解不香吗
by CSP_juruo @ 2023-08-13 10:49:59
@[louzezhong20130517](/user/989399) 通项公式没用，小数据不如暴力，大数据不如矩阵快速幂，别用通项公式
by Argvchs @ 2023-08-13 10:50:54
通项公式？
$$fib _ n = (\begin{bmatrix} 1 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} ^ {n - 2}) _ {1, 1}$$
大雾（）
by Iniaugoty @ 2023-08-13 11:18:22