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。

#include <bits/stdc++.h>
 
using namespace std;
 
#define REP(i,n)                for(int i(0); i <  (n); ++i)
#define rep(i,a,b)              for(int i(a); i <= (b); ++i)
#define dec(i,a,b)              for(int i(a); i >= (b); --i)
#define for_edge(i,x)           for(int i = H[x]; i; i = X[i])
 
#define LL      long long
#define ULL     unsigned long long
#define MP      make_pair
#define PB      push_back
#define FI      first
#define SE      second
#define INF     1 << 30
 
const int N     =    100000      +       10;
const int M     =    10000       +       10;
const int Q     =    1000        +       10;
const int A     =    30          +       1;
 
int a[N], b[N];
int n;
int ans;
 
int main(){
#ifndef ONLINE_JUDGE
	freopen("test.txt", "r", stdin);
	freopen("test.out", "w", stdout);
#endif
 
	scanf("%d", &n);
	rep(i, 1, n) scanf("%d", a + i);
	rep(i, 0, 1){
		bool flag = true;
		b[1] = i;
		rep(j, 2, n) b[j] = a[j - 1] - b[j - 1] - b[j - 2];
		rep(j, 2, n - 1) if (b[j] < 0 || b[j] > 1){ flag = false; break;}
		if (b[n] < 0 || b[n] > 1) flag = false;
		rep(j, 1, n) if (b[j - 1] + b[j] + b[j + 1] != a[j]){ flag = false; break;}
		//rep(j, 1, n) printf("%d ", b[j]); putchar(10);
 
		if (flag) ++ans;
	}
 
	printf("%d\n", ans);
 
	return 0;
 
}