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  • 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
    ", ans);
    48 
    49 
    50 
    51     return 0;
    52 
    53 }
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  • 原文地址:https://www.cnblogs.com/cxhscst2/p/6351834.html
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