"尚学堂杯"哈尔滨理工大学第七届程序设计竞赛 C.Collection Game(DP)

题目：

C.Collection Game

Time Limit: 1000 MS Memory Limit: 128000 K Total Submit: 841 (248 users) Total Accepted: 236 (165 users) Special Judge: No

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, 4000 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

Submit

Message

思路：
比赛的时候考虑的不周全，没有特判，其实就是一个菲波那切数列,哪几个点不能走就把哪几个点标记一下，然后在递推的时候判断一下就好。

代码：

#include <stdio.h>
#include <string.h>
#include <string>
#include <iostream>
#include <stack>
#include <cmath>
#include <queue>
#include <vector>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define inf 0x3f3f3f3f
#define N 20000+20
#define mod 10007
#define M 1000000+10
#define LL long long
using namespace std;
LL x,n,m;
LL num[10020];
bool flag[10020];
int main()
{

while(~scanf("%lld%lld",&n,&m))
{
mem(flag,0);
mem(num,0);
for(LL 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(LL 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
);
}
return 0;
}