我是这样做的,你和我的方法对比一下
```c
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;
int f[1001],p,res[1001],sav[1001];
void result_1(){
memset(sav,0,sizeof(sav));
for(register int i=1;i<=500;i+=1)
for(register int j=1;j<=500;j+=1)
sav[i+j-1]+=res[i]*f[j];
for(register int i=1;i<=500;i+=1){
sav[i+1]+=sav[i]/10;
sav[i]%=10;
}
memcpy(res,sav,sizeof(res));
}
void result_2(){
memset(sav,0,sizeof(sav));
for(register int i=1;i<=500;i+=1)
for(register int j=1;j<=500;j+=1)
sav[i+j-1]+=f[i]*f[j];
for(register int i=1;i<=500;i+=1){
sav[i+1]+=sav[i]/10;
sav[i]%=10;
}
memcpy(f,sav,sizeof(f));
}
int main(){
scanf("%d",&p);
printf("%d\n",(int)(log10(2)*p+1));
res[1]=1;
f[1]=2;
while(p!=0){
if(p%2==1)result_1();
p/=2;
result_2();
}
res[1]-=1;
for(register int i=500;i>=1;i-=1)
if(i!=500&&i%50==0)printf("\n%d",res[i]);
else printf("%d",res[i]);
return 0;
}
```
by 15167987933yy @ 2023-08-02 14:10:05
# 关于$\textcolor{red}{WA}$
啊题目说了:![](https://cdn.luogu.com.cn/upload/image_hosting/h93y8vom.png)
所以呢,我们要![](https://cdn.luogu.com.cn/upload/image_hosting/h93y8vom.png)
于是输出部分变成了
```cpp
for (int i=500;i>=1;i--)
{
printf("%d",gaojing[i]);
if(i%10==0)printf("\n");//换行
}
```
然后
$\textcolor{red}{0 \to 20 }$
# $关于大数据$
我们输入2000,得到了:
```
500
-19483298084596463898746277344711896086305533142593135616665
.....[后略]
```
答:$1.$后面疯狂向前面进位,前面炸了!$2.$而且,超过500位的部分就不知道前面有几位了!$3.$$\textcolor{brown}{TLE}$:总共乘310000次,每次乘500次,很容易就TLE!
1. 让```gaojing[500]```也向```gaojing[501]```进位就行了
2. 换底公式:
$$\log_b(N)=\frac{\log_a(N)}{\log_a(b)}$$
$\space\space\space\space\space$用换底公式把2的幂换成10的幂
3. 快速幂即可
后面我也真的不会了,~~还是看题解吧~~
by wanglexi @ 2023-09-06 19:53:49
@[wanglexi](/user/378403) 呃.....Markdown怎么坏了??
by wanglexi @ 2023-09-06 19:54:45
@[wanglexi](/user/378403) 没办法,在编辑区看都是好的
by wanglexi @ 2023-09-06 19:55:27
好吧,我确实挺蒟蒻的
by wanglexi @ 2023-09-06 19:59:23