HDU4403 A very hard Aoshu problem
搜索提，等号只有一个可以剪枝
1 #include <iostream> 2 #include <queue> 3 #include <cstring> 4 #include <cstdio> 5 using namespace std; 6 char seq[50]; 7 char str[50]; 8 int n,ans,eq; 9 bool ok(int len) 10 { 11 long long a,sum=0; 12 int i,flg=1; 13 queue<char> q; 14 q.push('+'); 15 seq[len]='\0'; 16 for(i=0; i<len; i++) 17 { 18 a=0; 19 while(i<=len&&seq[i]>='0'&&seq[i]<='9') a=a*10+seq[i]-'0',i++; 20 q.push(seq[i]); 21 if(flg&&q.front()=='+') 22 sum+=a; 23 else 24 flg=0,sum-=a; 25 q.pop(); 26 } 27 if(sum==0) return 1; 28 else return 0; 29 } 30 void dfs(int cur,int q) 31 { 32 if(eq>1) return; 33 if(q>n) 34 { 35 if(ok(cur)) ans++; 36 return; 37 } 38 seq[cur]=str[q]; 39 dfs(cur+1,q+1); 40 seq[cur]='+'; 41 seq[cur+1]=str[q]; 42 dfs(cur+2,q+1); 43 seq[cur]='='; 44 seq[cur+1]=str[q]; 45 eq++; 46 dfs(cur+2,q+1); 47 eq--; 48 } 49 int main() 50 { 51 while(cin>>str,str[0]!='E') 52 { 53 ans=0;eq=0; 54 seq[0]=str[0]; 55 n=strlen(str)-1; 56 dfs(1,1); 57 cout<<ans<<endl; 58 } 59 return 0; 60 }
