题意：http://poj.org/problem?id=3134
应该好理解
思路：
枚举层数（也就是ans）
dfs判断d到这个深度可不可以
+各种剪枝就能过
1 #define IOS ios_base::sync_with_stdio(0); cin.tie(0);
2 #include <cstdio>//sprintf islower isupper
3 #include <cstdlib>//malloc exit strcat itoa system("cls")
4 #include <iostream>//pair
5 #include <fstream>//freopen("C:\\Users\\13606\\Desktop\\Input.txt","r",stdin);
6 #include <bitset>
7 //#include <map>
8 //#include<unordered_map>
9 #include <vector>
10 #include <stack>
11 #include <set>
12 #include <string.h>//strstr substr strcat
13 #include <string>
14 #include <time.h>// srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
15 #include <cmath>
16 #include <deque>
17 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less
18 #include <vector>//emplace_back
19 //#include <math.h>
20 #include <cassert>
21 #include <iomanip>
22 //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor
23 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
24 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation
25 //******************
26 clock_t __START,__END;
27 double __TOTALTIME;
28 void _MS(){__START=clock();}
29 void _ME(){__END=clock();__TOTALTIME=(double)(__END-__START)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;}
30 //***********************
31 #define rint register int
32 #define fo(a,b,c) for(rint a=b;a<=c;++a)
33 #define fr(a,b,c) for(rint a=b;a>=c;--a)
34 #define mem(a,b) memset(a,b,sizeof(a))
35 #define pr printf
36 #define sc scanf
37 #define ls rt<<1
38 #define rs rt<<1|1
39 typedef pair<int,int> PII;
40 typedef vector<int> VI;
41 typedef unsigned long long ull;
42 typedef long long ll;
43 typedef double db;
44 const db E=2.718281828;
45 const db PI=acos(-1.0);
46 const ll INF=(1LL<<60);
47 const int inf=(1<<30);
48 const db ESP=1e-9;
49 const int mod=(int)1e9+7;
50 const int N=(int)1e6+10;
51
52 int n;
53 bool vis[N];
54 int a[N];
55 bool dfs(int now,int cnt,int top)
56 {
57 if(cnt==top)
58 {
59 if(now==n)return 1;
60 return 0;
61 }
62 if((now<<(top-cnt))<n)return 0;
63 if(now==n)return 1;
64 for(int i=1;i<=cnt;++i)
65 {
66 if(now-a[i]>0&&!vis[now-a[i]])
67 {
68 a[cnt+1]=now-a[i];
69 vis[now-a[i]]=1;
70 if(dfs(now-a[i],cnt+1,top))
71 {
72 vis[now-a[i]]=0;
73 return 1;
74 }
75 else
76 vis[now-a[i]]=0;
77 }
78 if(!vis[now+a[i]])
79 {
80 a[cnt+1]=now+a[i];
81 vis[now+a[i]]=1;
82 if(dfs(now+a[i],cnt+1,top))
83 {
84 vis[now+a[i]]=0;
85 return 1;
86 }
87 else
88 vis[now+a[i]]=0;
89 }
90 }
91 return 0;
92 }
93
94 int main()
95 {
96 while(sc("%d",&n),n!=0)
97 {
98 int ans=1;
99 vis[1]=1;
100 a[1]=1;
101 while(!dfs(1,1,ans))
102 {
103 ans++;
104 }
105 pr("%d\n",ans-1);
106 }
107 return 0;
108 }
109
110 /**************************************************************************************/
来源：https://www.cnblogs.com/--HPY-7m/p/12642979.html