#include <iostream>
//#include <fstream>
using namespace std;
int Double(int *s,int n=100)
{
for(int i=0;i<=n;i++)s[i]+=s[i];
for(int i=0;i<=n;i++) if(s[i]>9)s[i+1]+=s[i]/10,s[i]%=10;
return 0;
}
int add2(int *s,int n=100)
{
s[0]+=2;
for(int i=0;(i<=n)&&(s[i]>9);i++) s[i+1]+=s[i]/10,s[i]%=10;
return 0;
}
int hn(int num,int *s)
{
if(num==1) s[0]=2;
else hn(num-1,s),Double(s),add2(s);
return 0;
}
int main()
{
//ifstream cin("p1345.in",ios::in);
//ofstream cout("p1345.out",ios::out);
int num;
cin>>num;
int s[111]={0};
hn(num,s);
int l;
for(l=110;s[l]==0;l--);
for(int i=l;i>=0;i--)cout<<s[i];
//cin.close();
//cout.close();
return 0;
}

水题一个
#include<cstdio>
#include<cstring>
using namespace std;
int ans[1000],b[1000];
int main()
{
//freopen("hanoi.in","r",stdin);
//freopen("hanoi.out","w",stdout);
int n,la,i,j;
scanf("%d",&n);
memset(ans,0,sizeof(ans));
ans[1]=2;
la=1;
for(i=2;i<=n;i++)
{
memset(b,0,sizeof(b));
for(j=1;j<=la;j++)
{
b[j+1]=(b[j]+ans[j]*2)/10;
b[j]=(b[j]+ans[j]*2)%10;
}
b[1]+=2;
j=1;
while(b[j]>=10)
{
b[j+1]+=b[j]/10;
b[j]%=10;
j++;
}
la+=2;
while(b[la]==0)
la--;
for(j=la;j>=1;j--)
ans[j]=b[j];
}
for(i=la;i>=1;i--)
printf("%d",ans[i]);
printf("\n");
return 0;
}

###还要用到高精度
P1354Hanoi双塔问题Accepted
记录信息
评测状态 Accepted
题目 P1354 Hanoi双塔问题
递交时间 2015-08-15 21:20:13
代码语言 C++
评测机 VijosEx
消耗时间 81 ms
消耗内存 516 KiB
评测时间 2015-08-15 21:20:14
评测结果
编译成功

测试数据 #0: Accepted, time = 15 ms, mem = 516 KiB, score = 10
测试数据 #1: Accepted, time = 7 ms, mem = 516 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 516 KiB, score = 10
测试数据 #3: Accepted, time = 15 ms, mem = 516 KiB, score = 10
测试数据 #4: Accepted, time = 15 ms, mem = 516 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 516 KiB, score = 10
测试数据 #6: Accepted, time = 1 ms, mem = 516 KiB, score = 10
测试数据 #7: Accepted, time = 13 ms, mem = 516 KiB, score = 10
测试数据 #8: Accepted, time = 15 ms, mem = 516 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 512 KiB, score = 10
Accepted, time = 81 ms, mem = 516 KiB, score = 100
代码
#include <iostream>
#include <stdio.h>
using namespace std;
int ans[50];
long long work(int n)
{
if(n==1)return 2;
return (long long)2*work(n-1)+2;
}
int main()
{
int n;
scanf("%d",&n);
int maxnow=1;
for(int i=1;i<=n;i++)
{
ans[1]+=1;
int in=0;
int j;
for(j=1;j<=maxnow||in;j++)
{
ans[j]=2*ans[j]+in;
in=ans[j]/10;
ans[j]%=10;

}
if(j-1>maxnow)
maxnow=j-1;
}

for(int i=maxnow;i>0;i--)
printf("%d",ans[i]);
}

var
a:ARRAY[1..61] OF -2..20;
i,j,k,h,m,n:longint;
begin
readln(n);
a[1]:=2;
M:=1;
for J:=1 to n do
BEGIN
K:=0;
FOR I:=1 TO M DO
BEGIN
A[I]:=A[I]*2+K;
K:=A[I] DIV 10;
A[I]:=A[I] MOD 10;
END;
IF K>0 THEN BEGIN INC(M);
A[M]:=K;
END;
END;
I:=1;
DEC(A[I],2);
WHILE A[I]<0 DO
BEGIN
INC(A[I],10);
INC(I);
DEC(A[I],1);
END;
WHILE A[M]=0 DO DEC(M);
FOR I:=M DOWNTO 1 DO WRITE(A[I]);
END.

评测状态 Accepted
题目 P1354 Hanoi双塔问题
递交时间 2014-12-06 16:56:52
代码语言 Pascal
评测机 上海红茶馆
消耗时间 0 ms
消耗内存 816 KiB
评测时间 2014-12-06 16:56:52
评测结果
编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 812 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 816 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 812 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 816 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 816 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 816 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 812 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 816 KiB, score = 10
测试数据 #8: Accepted, time = 0 ms, mem = 812 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 812 KiB, score = 10
Accepted, time = 0 ms, mem = 816 KiB, score = 100
代码
var
st1,st2:string;
a,b,c,d:longint;
procedure jian;
var a,b,c,d,len1,len2:longint;
s2:string;
begin
len1:=length(st1);
s2:='2';
len2:=length(s2);
a:=0;
while len2<>0 do
begin
b:=ord(st1[len1])-48;
c:=ord(st2[len2])-48;
b:=b-c-a;
if b<0 then
begin
b:=b+10;
a:=1;
end
else
a:=0;
st1[len1]:=chr(b+48);
dec(len1);
dec(len2);
end;
a:=1;
while st1[a]='0' do inc(a);
if a=length(st1)+1 then
begin
write('0');
exit;
end;
for b:=a to length(st1) do
begin
write(st1[b]);
c:=1;
end;
end;
procedure cheng;
var
a,b:array[1..250] of integer;
c:array[1..500] of integer;
i,j,len,len1,len2,t:integer;
fh1,fh2:char;
begin
len1:=length(st1);
len2:=length(st2);
fillchar(c,sizeof(a),0);
for i:=1 to len1 do
a[i]:=ord(st1[len1-i+1])-48;
for i:=1 to len2 do
b[i]:=ord(st2[len2-i+1])-48;
for i:=1 to len1 do
begin
t:=i;
for j:=1 to len2 do
begin
c[t]:=c[t]+a[i]*b[j];
c[t+1]:=c[t] div 10+c[t+1];
c[t]:=c[t] mod 10;
inc(t);
end;
end;
while c[t]=0 do t:=t-1; st1:='';
for i:=t downto 1 do st1:=st1+chr(c[i]+48);
end;

begin
st1:='2';
st2:=st1;
readln(a);
for b:=1 to a do cheng;
jian;
end.

数学自己解吧。一个数列的通项公式而已。
program hanoi;
var n:integer;
a:array[1..10000] of longint;
i,x,m,c,line,num:longint;
op:boolean;
begin
//assign(input,'hanoi.in'); reset(input);
//assign(output,'hanoi.out'); rewrite(output);
op:=false;
read(n);
a[1]:=1;
m:=1;
x:=0;
line:=1;
if n=1 then
write('2')
else
begin
for i:=1 to n do
begin
for num:=1 to line do
a[num]:=a[num]*2;

for num:=1 to line do
if a[num]>=10 then
begin x:=a[num] div 10; a[num]:=a[num] mod 10; a[num+1]:=a[num+1]+x; end;

if a[num+1]<>0 then
line:=line+1;
end;

for c:=10000 downto 1 do
begin
if (a[c]<>0) and (op=false) then
begin op:=true; line:=c; end;
if op=true then
break;
end;

a[1]:=a[1]-1;
while a[m]=-1 do
begin
a[m]:=9;
m:=m+1;
a[m]:=a[m]-1;
end;

m:=0;
for i:=1 to line do
a[i]:=a[i]*2;

for i:=1 to line do
if a[i]>=10 then
begin x:=a[i] div 10; a[i]:=a[i] mod 10; a[i+1]:=a[i+1]+x; end;

if a[i+1]<>0 then
line:=line+1;

for i:=line downto 1 do
write(a[i]);

//close(input); close(output);
end;
readln;
end.

其实是道python题
print 2**input()*2-2

#include <iostream>

#include <string>
using namespace std;
int main ()
{
int n,l=1,a[1001]={0};
cin>>n;
n++;
a[1]=1;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=100;j++)
a[j]*=2;
for(int j=1;j<=100;j++)
{
if(a[j]>=10)
{
a[j+1]+=a[j]/10;
a[j]%=10;
}
}
if(i==n-1)a[1]--;
}
for(int i=1;i<=1000;i++)
if(a[i]!=0)l=max(i,l);
for(int i=l;i>=1;i--)
cout<<a[i];
return 0;
}

#include<cstdio>
#include<cstdlib>
int ans[1000010]={1,1},n;
int main()
{
scanf("%d",&n);
for(int t=1;t<=n+1;t++)
{
ans[1]*=2;
for(int i=1;i<=ans[0];i++)
ans[i+1]*=2,ans[i+1]+=ans[i]/10,ans[i]%=10;
ans[ans[0]+1]?ans[0]++:1;
}
ans[1]-=2;
for(int i=2;i<=ans[0];i++)
{
if(ans[i-1]<0) ans[i-1]+=10,ans[i]--;
else break;
}
for(int i=ans[0];i;i--)
putchar(ans[i]+48);
puts("");
system("pause");
return 0;
}

#include<cstdio>
#include<cstdlib>
int ans[1000010]={1,1},n;
int main()
{
scanf("%d",&n);
for(int t=1;t<=n+1;t++)
{
ans[1]*=2;
for(int i=1;i<=ans[0];i++)
ans[i+1]*=2,ans[i+1]+=ans[i]/10,ans[i]%=10;
ans[ans[0]+1]?ans[0]++:1;
}
ans[1]-=2;
for(int i=2;i<=ans[0];i++)
{
if(ans[i-1]<0) ans[i-1]+=10,ans[i]--;
else break;
}
for(int i=ans[0];i;i--)
putchar(ans[i]+48);
puts("");
system("pause");
return 0;
}

function f(x:longint):longint;
begin
if x=1 then f:=1 else
f:=f(x-1)*2+1
end;
var n:longint;
begin
read(n);
write(2*f(n));
end.

题目分析：
本题用记忆化递归，然后再加上高精度就可以过了。
设F[n]表示有2*n个圆盘时完成题目所说的任务所需的最少移动次数，则
F[n]:=2*f[n-1]+2；
type data=array[0..1000] of longint;
var
n,i:longint;
two,zero:data;
f:array[0..200] of data;
function add(a,b:data):data;
var i,len:longint;
c:data;
begin
fillchar(c,sizeof(c),0);
if a[0]>b[0] then len:=a[0] else len:=b[0];
for i:=1 to len do
begin
c[i]:=a[i]+b[i]+c[i];
c[i+1]:=c[i] div 10+c[i+1];
c[i]:=c[i] mod 10;
end;
if c[len+1]>0 then inc(len);
c[0]:=len;
exit(c);
end;
procedure sub(n:longint);
begin
if not((f[n,0]=1)and(f[n,1]=0)) then exit;
sub(n-1);
f[n]:=add(add(f[n-1],f[n-1]),two);
end;
begin
readln(n);
zero[0]:=1; zero[1]:=0;
two[0]:=1; two[1]:=2;
for i:=0 to n do f[i]:=zero;
f[1]:=two;
sub(n);
for i:=f[n,0] downto 1 do write(f[n,i]);
end.