Hrbust 2327 Collection Game（思维）

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原创最新推荐文章于 2026-05-18 10:15:17 发布 442 阅读 AI 写同款· GEO 优化 ›

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Collection Game

Time Limit: 1000 MS

Memory Limit: 128000 K

Total Submit: 92(39 users)

Total Accepted: 40(35 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

题意：上楼梯，起点在一阶，每次可走一阶两阶或三阶，但有一些台阶坏了不能走，问到第n阶楼梯有多少种走法。

思路：类似于斐波那契数列，a[i] = a[i-1] + a[i-2] + a[i-3] 。

#include<bits/stdc++.h>
using namespace std;
#define Mod 10007
const int N = 10000 + 10 ;
int a[N],b[N];
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)==2)
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(int i=0;i<m;i++)
        {
            int x;
            scanf("%d",&x);
            b[x]=1;
        }
        a[2]=b[2]?0:1;
        a[3]=b[3]?0:a[2]+1;
        a[4]=b[4]?0:a[2]+a[3]+1;
        for(int i=5;i<=n;i++)
        {
            if(b[i])continue ;
            if(!b[i-1])
                a[i]+=a[i-1];
            if(!b[i-2])
                a[i]+=a[i-2];
            if(!b[i-3])
                a[i]+=a[i-3];
            a[i]%=Mod;
        }
        printf("%d\n",a[n]);
    }
}

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