UVa-12107 Digit Puzzle

“借鉴”了别人的代码，优化十分巧妙——只需递归前两个数，就可知道能不能凑一个第三个数使这种情况成立。
  1 #include <bits/stdc++.h>
  2 #define _for(i,a,b) for(int i = (a);i < (b);i ++)
  3 #define pb push_back
  4 using namespace std;
  5 //-2 stands for the blank ;-1 stands for the *
  6 vector<int> p[3];
  7 
  8 bool ok()
  9 {
 10     int d[2] {0};
 11     _for(i,0,2)
 12     _for(j,0,p[i].size())
 13     if(p[i][j]==-1)
 14         return false;
 15     else
 16         d[i] = d[i]*10+p[i][j];
 17         
 18     int num = d[0]*d[1];
 19     int str[39];
 20     
 21     _for(i, 0, p[2].size())
 22     {
 23         str[p[2].size()-i-1] = num % 10;
 24         num /= 10;
 25     }
 26     
 27     if(num != 0 || str[0] == 0) return 0;
 28     _for(i, 0, p[2].size())
 29         if(p[2][i] != -1 && str[i] != p[2][i])
 30             return 0;
 31             
 32     return 1;
 33 }
 34 
 35 int dfs1(int i,int j)
 36 {
 37     if(i==2)
 38         return ok();
 39     int rnt = 0;
 40     if(p[i][j]==-1)
 41     {
 42         _for(k,0,10)
 43         {
 44             if(k==0&&j==0)
 45                 continue;
 46             p[i][j] = k;
 47             if(j==p[i].size()-1)
 48             {
 49                 rnt += dfs1(i+1,0);
 50                 p[i][j] = -1;
 51                 if(rnt>1) return 2;
 52             }
 53             else
 54             {
 55                 rnt += dfs1(i,j+1);
 56                 p[i][j] = -1;
 57                 if(rnt>1) return 2;
 58             }
 59         }
 60     }
 61     else
 62     {
 63         if(j==p[i].size()-1)
 64         {
 65             rnt += dfs1(i+1,0);
 66             if(rnt>1) return 2;
 67         }
 68         else
 69         {
 70             rnt += dfs1(i,j+1);
 71             if(rnt>1) return 2;
 72         }
 73     }
 74     return rnt;
 75 }
 76 
 77 int maxd;
 78 const char* word = "*0123456789";
 79 bool dfs2(int i,int j,int d)
 80 {
 81     if(d==maxd) return (dfs1(0,0)==1);
 82     if(i==3) return false;
 83 
 84     int ti, tj;
 85     if(j == p[i].size() - 1){ti = i + 1;tj = 0;}
 86     else{ti = i;tj = j + 1;}
 87     
 88     char t;
 89     if(p[i][j]==-1)
 90         t = '*';
 91     else
 92         t = p[i][j] + '0';
 93     _for(k, 0, 11)
 94     {
 95         if(word[k] == '0' && j == 0) continue;
 96         if(t == word[k])
 97         {
 98             if(dfs2(ti, tj, d))
 99                 return true;
100         }
101         else
102         {
103             if(k==0)
104             {
105                 p[i][j] = -1;
106                 if(dfs2(ti, tj, d + 1)) return true;
107             }
108             else
109             {
110                 p[i][j] = word[k]-'0';
111                 if(dfs2(ti, tj, d + 1)) return true;
112             }
113             if(t=='*')
114                 p[i][j] = -1;
115             else
116                 p[i][j] = t-'0';
117         }
118     }
119     return false;
120 }
121 
122 void ID()
123 {
124     for(maxd=0; ; maxd ++)
125         if(dfs2(0,0,0)) return ;
126 }
127 
128 int kase = 1;
129 void output()
130 {
131     printf("Case %d: ",kase++);
132     _for(i,0,3)
133     {
134         _for(j,0,p[i].size())
135         {
136             if(p[i][j]==-1)
137                 cout << '*';
138             else
139                 cout << p[i][j];
140         }
141         if(i!=2)
142             cout << " ";
143     }
144     cout << endl;
145 }
146 
147 int main()
148 {
149     string str;
150     while(getline(cin,str)&&str[0]!='0')
151     {
152         _for(i,0,3)
153         p[i].clear();
154         int pEnd = 0;
155         _for(i,0,str.size())
156         {
157             if(str[i]=='*')
158                 p[pEnd].pb(-1);
159             else if(str[i]==' ')
160                 pEnd ++;
161             else
162                 p[pEnd].pb(str[i]-'0');
163         }
164         ID();
165         output();
166     }
167     return 0;
168 }