（尚學杯）C.Collection Game

Description

POI and POJ are pair of sisters, one is a master in “Kantai Collection”, another is an excellent competitor in ACM programming competition. One day, POI wants to visit POJ, and the pace that between their homes is made of square bricks. We can hypothesis that POI’s house is located in the NO.1 brick and POJ’s house is located in the NO.n brick. For POI, there are three ways she can choose to move in every step, go ahead one or two or three bricks. But some bricks are broken that couldn’t be touched. So, how many ways can POI arrive at POJ’s house?

Input

There are multiple cases.

In each case, the first line contains two integers N(1<=N<=10000) and M (1<=M<=100), respectively represent the sum of bricks, and broke bricks. Then, there are M number in the next line, the K-th number a[k](2<=a[k]<=n-1) means the brick at position a[k] was broke.

Output

Please output your answer P after mod 10007 because there are too many ways.

Sample Input

5 1

3

Sample Output

3

這題是一道斐波那契數列題，只是需要判斷一下有什麼點不能走

比賽的時候忘記特判了（還是有點緊張），LOL

代碼：

#include <stdio.h>
#include <string.h>
#include <iostream>
#define mod 10007
using namespace std;
long long x,n,m;
long long num[10020];
bool flag[10020];
int main()
{
    while(~scanf("%lld%lld",&n,&m))
    {
        memset(flag,0,sizeof(flag));
        memset(num,0,sizeof(num));
        for(long long i=0; i<m; i++)
        {
            scanf("%lld",&x);
            flag[x]=1;
        }
        num[1]=1;
        if(!flag[2]) num[2]=1;
        if(!flag[3]) num[3]=num[2]+1;
        for(long long i=4; i<=n; i++)
        {
            if(flag[i])
                num[i]=0;
            else
                num[i]=(num[i-1]+num[i-2]+num[i-3])%mod;
        }
        printf("%lld\n",num[n]);
    }
    return 0;
}