題面

原題

Solution

我們考慮第一個雷如果確定了，顯然後面每一個地方是否有雷都確定了，於是考慮2次遞推。

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#include<queue>
#define ll long long
#define file(a) freopen(a".in","r",stdin)//;freopen(a".out","w",stdout)
using namespace std;
inline int gi(){
    int sum=0,f=1;char ch=getchar();
    while(ch>'9' || ch<'0'){if(ch=='-')f=-f;ch=getchar();}
    while(ch>='0' && ch<='9'){sum=(sum<<3)+(sum<<1)+ch-'0';ch=getchar();}
    return f*sum;
}
inline ll gl(){
    ll sum=0,f=1;char ch=getchar();
    while(ch>'9' || ch<'0'){if(ch=='-')f=-f;ch=getchar();}
    while(ch>='0' && ch<='9'){sum=(sum<<3)+(sum<<1)+ch-'0';ch=getchar();}
    return f*sum;
}
int a[500010],flag[500010],ans1,ans2;
int main(){
#ifndef ONLINE_JUDGE
    file("example");
#endif
    int i,j,n,m,k;
    n=gi();for(i=1;i<=n;i++)a[i]=gi();
    flag[1]=0;ans1=1,ans2=1;
    for(i=2;i<=n+1;i++){
        flag[i]=a[i-1]-flag[i-1]-flag[i-2];
        if(flag[i]<0 || flag[i]>1){ans1=0;break;}
    }
    if(flag[n+1])ans1=0;
    flag[1]=1;
    for(i=2;i<=n+1;i++){
        flag[i]=a[i-1]-flag[i-1]-flag[i-2];
        if(flag[i]<0 || flag[i]>1){ans2=0;break;}
    }
    if(flag[n+1])ans2=0;
    printf("%d\n",ans1+ans2);
    return 0;
}