嗯，这算法应该对的，就是慢了点。

/*---------------------------------------------------------------------------
File name: NKU1021-[NKPC2]?S???.c
Author: HJin (email: fish_sea_bird [at] yahoo [dot] com )
Created on: 8/12/2007 21:29:53
Environment: Windows XP Professional SP2 English +
Visual Studio 2005 v8.0.50727.762

Modification history:
===========================================================================

I think I discussed this problem with Leeco before. Bottom of line is:

write out the binary representation of n.

Following is my code submitted to Nankai online judge.
*/

long long b[] = {1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987, 22876792454961, 68630377364883, 205891132094649, 617673396283947};

int main()
{
int k;
unsigned n;

while( scanf("%u", &n) != -1 )
{
--n;
k=0;

while(n)
{
if(n&1)
printf("%lld\n", b[k]);

++k;
n >>= 1;
}
}

return 0;
}

这个算法应该没有错啊

这个是我的 算法好像一样吧
#include <stdio.h>
#include<math.h>
int main()
{
int k ,n,i,j,m,t,L,y;
int a[1000]={0};
a[0]=1;
while(scanf("%d%d",&k,&n)!=EOF){
j=1,t=0,L=1,y=2;
for(m=0;m<n;m++)
{
if(j+1==y){
y=y*2;
t=pow(k,L++);
a[j++]=t;
i=0;
continue;
}
a[j++]=t+a[i++];
}
printf("%d",a[n-1]);

}
}

//回复11楼：

#include<iostream>
#include<cmath>
using namespace std;

int main()
{
short x=0,s[2002];
short i,j,n,k;
cin >> k >> n;
i=1;//原来的0改为1就成功啦
while(i<=n)
{
s[i]=(short)pow(k,x);
x++;
for(j=1;j<i;j++)
{
if(i+j>=2002)return -1;
s[i+j]=s[i]+s[j];
}
i=i+j;
}
cout<<s[n]<<endl;
return 0;
}