HDU - 5573 Binary Tree-蒲公英云

HDU - 5573 Binary Tree

题目：点击打开链接题意：给你一个n和k和一颗完全二叉树（权值1 2 3 4 5 6 …）,让你从1开始一直到第k层选择一条路径，路径上的每个点可以选择1或(-1)，让你输出一种构造方案恰好到第n层能得到n。分析：很有意思的一道题，记录一下，首先明确由1.2.4…2k1.2.4…2k可以构造出所有小于2k+12k+1的数，那么实际上只要走2的幂次即最左边的结点即可。那么实际上这个过程就类似整数的二进制表示，1表示加，0表示减，如果该位为0，那么意味着在2k+1−12k+1−1的基础上减去了二倍改为代表的数，所以我们求出n与2k+1−12k+1−1的差，将差的一半的二进制中为1的位置为0即可~~ 注意偶数的时候，差为奇数，那么我们将差多加个一，让最后一步走右边的结点，即加一即可。参考博客https://blog.csdn.net/yukizzz/article/details/51264836。代码：

///#pragma comment(linker, "/STACK:102400000,102400000")
///#include<unordered_map>
///#include<unordered_set>
#include<algorithm>
#include<iostream>
#include<fstream>
#include<complex>
#include<cstdlib>
#include<cstring>
#include<cassert>
#include<iomanip>
#include<string>
#include<cstdio>
#include<bitset>
#include<vector>
#include<cctype>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<deque>
#include<list>
#include<set>
#include<map>
using namespace std;
#define pt(a) cout<<a<<endl
#define debug test
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pii pair<int,int>
#define fi first
#define se second
#define ll long long
#define ull unsigned long long
#define pb push_back
#define mp make_pair
#define inf 0x3f3f3f3f
#define eps 1e-10
#define PI acos(-1.0)
const ll mod = 1e9+7;
const int N = 1e6+10;
ll qp(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
int to[4][2]={
   {-1,0},{1,0},{0,-1},{0,1}};
int a[N];
int main ()
{
    int t;
    cin>>t;
    int cnt = 0;
    while(t--){
        int n, k;
        cin>>n>>k;
        cnt++;
        cout<<"Case #"<<cnt<<":"<<endl;
        ll sum = 1<<(k+1) - 1;
        int f = 0;
        if(n % 2 == 0){
            f = 1;
            n--;
        }
        ll res = (sum - n) / 2;
        for(int i = 0; i < k - 1; i++){
            ll tmp = 1<<i;
            if(res&1)  cout<<tmp<<" -"<<endl;
            else cout<<tmp<<" +"<<endl;
            res = res>>1;
        }
        ll tmp = 1<< (k - 1);
        if(f) cout<<tmp + 1<<" +"<<endl;
        else cout<<tmp<<" +"<<endl;
    }
    return 0;
}