POJ - 2248 迭代加深

Posted 2020-10-24 The Azure Arbitrator

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