Problem:
Score : 300 points

Problem Statement

There is a staircase with N steps. Takahashi is now standing at the foot of the stairs, that is, on the 0-th step.
He can climb up one or two steps at a time.

However, the treads of the a_1-th, a_2-th, a_3-th, \ldots, a_M-th steps are broken, so it is dangerous to set foot on those steps.

How many are there to climb up to the top step, that is, the N-th step, without setting foot on the broken steps?
Find the count modulo 1\ 000\ 000\ 007.

Constraints

1 \leq N \leq 10^5

0 \leq M \leq N-1

1 \leq a_1 < a_2 < ... < a_M \leq N-1

Input

Input is given from Standard Input in the following format:

N M
a_1
a_2
.
.
.
a_M

Output

Print the number of ways to climb up the stairs under the condition, modulo 1\ 000\ 000\ 007.

Sample Input 1

6 1
3

Sample Output 1

4

There are four ways to climb up the stairs, as follows:

0 \to 1 \to 2 \to 4 \to 5 \to 6

0 \to 1 \to 2 \to 4 \to 6

0 \to 2 \to 4 \to 5 \to 6

0 \to 2 \to 4 \to 6

Sample Input 2

10 2
4
5

Sample Output 2

0

There may be no way to climb up the stairs without setting foot on the broken steps.

Sample Input 3

100 5
1
23
45
67
89

Sample Output 3

608200469

Be sure to print the count modulo 1\ 000\ 000\ 007.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#define N 100000
#define MOD 1000000007

static int memo[N+1] = {};

int calc(int n){
    int n1, n2;
    if(n == 0){
        return 1;
    }else if(n == 1){
        memo[1] = 1;
        return 1;
    }else if(n == 2){
        memo[2] = 2;
        return 2;
    }else{
        if(memo[n-2] == 0){
            n2 = calc(n-2);
            memo[n-2] = n2;
        }else{
            n2 = memo[n-2];
        }
        
        if(memo[n-1] == 0){
            n1 = calc(n-1);
            memo[n-1] = n1;
        }else{
            n1 = memo[n-1];
        }

        if(memo[n] == 0){
            memo[n] = n2 + n1;
        }
        return n2 + n1;
    }
}

int main(void){
    int n,m;
    int broken[N] = {};
    
    int i,j,k,b;
    int stages;
    long long int count = 1,countp;
    
    scanf("%d %d", &n, &m);
    
    for(i = 0; i < m; i++){
        scanf("%d", &broken[i]);
    }
    
    k = broken[0];
    for(i = 1; i < m; i++){
        if(broken[i] - k == 1){
            count = 0;
        }else{
            k = broken[i];
        }
    }
    
    if(count){
        if(m == 0){
            count = calc(n) % MOD;
        }else if(m == 1){
            count = calc(broken[0] - 1) % MOD;
            count = count * (calc(n-broken[0]-1) % MOD);
        }else{
            count = calc(broken[0] - 1) & MOD;
            for(i = 1; i < m; i++){
                count = count * (calc(broken[i] - broken[i-1] - 2) % MOD);
                count = count % MOD;
            }
            count = count * (calc(n-broken[m-1]-1) % MOD);
        }
    }
    
    count = count % MOD;
    
    printf("%lld", count);
    
	return 0;
}