2017广东工业大学程序设计竞赛决赛C.Collection Game（dp）

C.Collection GameTime Limit: 1000 MSMemory Limit: 128000 KTotal Submit: 841(248 users)Total Accepted: 236(165 users)Special Judge: NoDescriptionPOI 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.

OutputPlease output your answer P after mod 10007 because there are too many ways.Sample Input

5 1

3

Sample Output3

所有点都能走的情况是一个变形的斐波那契数列f[i]=f[i-1]+f[i-2]+f[i-3]，标记一下不能走的点，让其值为0，从而后面递推出的值都没有了从该点推出的值，即走过这点的所有情况都被排除

#include <iostream>#include<string>#include<string.h>#include<stdio.h>#include<stdlib.h>#include<queue>#include<math.h>#include<algorithm>#define ll long long#define mem(a,b) memset(a,b,sizeof(a))#define inf 0x3f3f3f3f#define maxn 100000+5using namespace std;ll f[10005],vis[10005];int main(){    ll n,m,x;    while(~scanf("%lld%lld",&n,&m))    {        mem(vis,0);        for(int i=1;i<=m;i++)        {            scanf("%lld",&x);            vis[x]=1;        }        f[1]=1;        if(!vis[2])f[2]=1;        if(!vis[3])f[3]=f[2]+1;        for(int i=4;i<=n;i++)        {            if(!vis[i])            f[i]=(f[i-1]+f[i-2]+f[i-3])%10007;            else            f[i]=0;        }        printf("%lld\n",f[n]);    }}