uva 529 - Addition Chains

Addition Chains

An addition chain for n is an integer sequence  with the following four properties:

• a₀ = 1
• a_m = n
• a₀1

2<...m-1m

For each k ( ) there exist two (not neccessarily different) integers i and j ( ) with a_k=a_i +a_j

You are given an integer n. Your job is to construct an addition chain for n with minimal length. If there is more than one such sequence, any one is acceptable.
For example, <1> and <1> are both valid solutions when you are asked for an addition chain for 5.

Input Specification 

The input file will contain one or more test cases. Each test case consists of one line containing one integer n (). Input is terminated by a value of zero (0) for n.

Output Specification 

For each test case, print one line containing the required integer sequence. Separate the numbers by one blank.

Hint: The problem is a little time-critical, so use proper break conditions where necessary to reduce the search space.

Sample Input 

5
7
12
15
77
0

Sample Output 

1 2 4 5
1 2 4 6 7
1 2 4 8 12
1 2 4 5 10 15
1 2 4 8 9 17 34 68 77

Iterative DFS + pruning 

#include<stdio.h>
#include<cstring>
#include<cmath>
#include<string>
#include<ctime>
#include<vector>
#define REP(i, b, n) for (int i = b; i < n; i++)
#define rep(i, n) REP(i, 0, n)
#define DBG 0
using namespace std;

int N,CM;
int cnt[10001];
bool has[10001];
vector<int>mv,cv;
void print(){
    bool ll=0;
    rep(i,mv.size()){if(ll)printf(" ");ll=1;printf("%d",mv[i]);}
    printf("\n");
}
bool DFS(int cur){
    if(DBG)printf("dfs cur %d size %d\n",cur, cv.size());
    if(cur==N){mv=cv;return 1;}
    else{
        int c,bc,k,j;
        has[cur]=1;
        rep(i,cv.size()){
            c=cur+cv[i];
            for(k=c,j=cv.size()+2;j<CM+1;k*=2,j++);
            if(k<N)continue;
            if(c<=N&&!has[c]&&cv.size()+1<=cnt[c]){
                cnt[c]=cv.size()+1;
                cv.push_back(c);
                if(DFS(c))return 1;
                cv.pop_back();
            }
        }
        has[cur]=0;
        return 0;
    }
    
}
void init(){
    memset(cnt,0x7f,sizeof(cnt));
}
int getNum(int a){
    return int(ceil(log(a)/log(2)));
}
void ans(){
    memset(has,0,sizeof(has));
    cv.clear(),cv.push_back(1),CM=18,has[0]=1;
    if(DBG)printf("cur %d\n",N);
    int n=getNum(N);
    REP(i,n,19){
        mv.clear(),CM=i;
        DFS(1);
        if(mv.size())break;
    }
    print();
}
int main(){
    int a;
    init();
    while(scanf("%d",&N)==1){
        if(N==0)break;
        ans();
    }
    //int t=clock();
    //printf ("%f seconds.\n",((float)t)/CLOCKS_PER_SEC);
}