"尚学堂杯"哈尔滨理工大学第七届程序设计竞赛 C.Collection Game【递推】

Collection Game

Time Limit: 1000 MS Memory Limit: 128000 K Total Submit: 41(21 users) Total Accepted: 22(20 users) Rating:  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, 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

Source

"尚学堂杯"哈尔滨理工大学第七届程序设计竞赛

题目大意：

一共给你N个台阶，每次可以走1步，2步，3步，但是有m个台阶不能走，问走到n号台阶有多少种方法。

思路：

dp【i】=dp【i-1】+dp【i-2】+dp【i-3】；

if（位子i是不能走的）continue掉就可以了；

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
#define mod 10007
int dp[10050];
int vis[10050];
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
memset(vis,0,sizeof(vis));
memset(dp,0,sizeof(dp));
while(m--)
{
int x;scanf("%d",&x);
vis[x]=1;
}
dp[1]=1;
for(int i=2;i<=n;i++)
{
if(vis[i]==1)continue;
if(i-1>=0)dp[i]+=dp[i-1];
if(i-2>=0)dp[i]+=dp[i-2];
if(i-3>=0)dp[i]+=dp[i-3];
dp[i]%=mod;
}
printf("%d\n",dp
);
}
}