枚举ak的值
反向枚举使ak尽快到达最短链的n
/*H E A D*/
int n,m,a[23333],dep;
bool dfs(int x){
if(a[x-1]>n||a[x-1]<=a[x-2])return 0;
if(x>dep){
if(a[x-1]==n) return 1;
else return 0;
}
// rep(i,1,x-1){
// rep(j,1,x-1){
rrep(i,x-1,1){
rrep(j,x-1,i){
a[x]=a[i]+a[j];
if(dfs(x+1))return 1;
}
}
return 0;
}
int main(){
IOS;
while(cin>>n){
if(n==0)break;
if(n==1){
cout<<1<<endl;
continue;
}
a[1]=1;
rep(depth,2,oo){
dep=depth;
if(dfs(2)){
rep(i,1,depth-1) cout<<a[i]<<" ";
cout<<a[depth]<<endl;
break;
}
}
}
return 0;
}
UPD:UVa上的TLE了啊,待我摸鱼过后再改改