BZOJ1088(SCOI2005)

  枚举第一行第一个格子的状态(有雷或者无雷,0或1),然后根据第一个格子推出后面所有格子的状态。推出之后判断解是否可行即可。

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 #define REP(i,n)                for(int i(0); i <  (n); ++i)
 6 #define rep(i,a,b)              for(int i(a); i <= (b); ++i)
 7 #define dec(i,a,b)              for(int i(a); i >= (b); --i)
 8 #define for_edge(i,x)           for(int i = H[x]; i; i = X[i])
 9 
10 #define LL      long long
11 #define ULL     unsigned long long
12 #define MP      make_pair
13 #define PB      push_back
14 #define FI      first
15 #define SE      second
16 #define INF     1 << 30
17 
18 const int N     =    100000      +       10;
19 const int M     =    10000       +       10;
20 const int Q     =    1000        +       10;
21 const int A     =    30          +       1;
22 
23 int a[N], b[N];
24 int n;
25 int ans;
26 
27 int main(){
28 #ifndef ONLINE_JUDGE
29     freopen("test.txt", "r", stdin);
30     freopen("test.out", "w", stdout);
31 #endif
32 
33     scanf("%d", &n);
34     rep(i, 1, n) scanf("%d", a + i);
35     rep(i, 0, 1){
36         bool flag = true;
37         b[1] = i;
38         rep(j, 2, n) b[j] = a[j - 1] - b[j - 1] - b[j - 2];
39         rep(j, 2, n - 1) if (b[j] < 0 || b[j] > 1){ flag = false; break;}
40         if (b[n] < 0 || b[n] > 1) flag = false;
41         rep(j, 1, n) if (b[j - 1] + b[j] + b[j + 1] != a[j]){ flag = false; break;}
42         //rep(j, 1, n) printf("%d ", b[j]); putchar(10);
43 
44         if (flag) ++ans;
45     }
46 
47     printf("%d\n", ans);
48 
49 
50 
51     return 0;
52 
53 }

 

posted @ 2017-01-26 19:31  cxhscst2  阅读(...)  评论(... 编辑 收藏