NYOJ 491 幸运三角形(bitset)

 1 #include<iostream>
 2 #include<cmath>
 3 #include<cstdio>
 4 #include<bitset>
 5 using namespace std;
 6 int a[22],n;
 7 int b[] = {0,0,0,4,6,0,0,12,40,0,0,171,410,0,0,1896,5160,0,0,32757};
 8 
 9 void slove()
10 {
11     freopen("out.txt","w",stdout);
12     int i, j, s1, s2, n,sum, T;
13     n = 1;
14     while(n < 20)
15     {
16         T = pow(2.0, n);
17         sum = 0;
18         while(T--)
19         {
20             bitset<20> b(T);
21             s1 = s2 = 0;
22             for(i=0; i < n; ++i)
23             {
24                 a[i] = b[i];
25                 if(a[i]) ++s1;
26                 else     ++s2;
27             }
28             for(j = 0; j < n; ++j)
29             {
30                 for(i = 0; i < n-j-1; ++i)
31                 {
32                     if(a[i] == a[i+1])
33                     {    a[i] = 1;    ++s1;  }
34                     else
35                     {    a[i] = 0;    ++s2;  }
36                 }
37             }
38             if(s1 == s2)
39                 ++sum;
40         }
41         b[n++] = sum;
42         cout<<","<<b[n-1];
43     }
44     cout<<endl;
45 }
46 
47 int main()
48 {
49 //    slove();
50     while(~scanf("%d",&n))
51         printf("%d\n",b[n]);
52     return 0;
53 }

先暴力解开答案，然后存入数组打表即可。。。做的时候突然想到可以用bitset<>...,很方便的哦。。。