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 ",ans-1); 106 } 107 return 0; 108 } 109 110 /**************************************************************************************/