XTU 1233 Coins(DP)

时间：2022-10-24 10:58:13

题意: n个硬币摆成一排，问有连续m个正面朝上的硬币的序列种数。

很明显的DP题。定义状态dp[i][1]表示前i个硬币满足条件的序列种数。dp[i][0]表示前i个硬币不满足条件的序列种数。

那么显然有dp[i][1]=dp[i-1][1]*2+dp[i-1-m][0].

如果前i-1个硬币满足条件，那么第i个硬币无论怎么样都满足条件。如果前i-1-m个硬币不满足条件，那么只需要再添加m个正面朝上的硬币即可。

dp[i][0]=2^i-dp[i][1].

于是最后的答案就是dp[n][1].

# include <cstdio>

# include <cstring>

# include <cstdlib>

# include <iostream>

# include <vector>

# include <queue>

# include <stack>

# include <map>

# include <set>

# include <cmath>

# include <algorithm>

using namespace std;

# define lowbit(x) ((x)&(-x))

# define pi acos(-1.0)

# define eps 1e-

# define MOD

# define INF

# define mem(a,b) memset(a,b,sizeof(a))

# define FOR(i,a,n) for(int i=a; i<=n; ++i)

# define FO(i,a,n) for(int i=a; i<n; ++i)

# define bug puts("H");

# define lch p<<,l,mid

# define rch p<<|,mid+,r

# define mp make_pair

# define pb push_back

typedef pair<int,int> PII;

typedef vector<int> VI;

# pragma comment(linker, "/STACK:1024000000,1024000000")

typedef long long LL;

int Scan() {

    int x=,f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

void Out(int a) {

    if(a<) {putchar('-'); a=-a;}

    if(a>=) Out(a/);

    putchar(a%+'');

}

const int N=;

//Code begin...

LL dp[N][], p[N];

void init(){p[]=; FO(i,,N) p[i]=p[i-]*%MOD;}

int main ()

{

    int T, n, m;

    scanf("%d",&T); init();

    while (T--) {

        scanf("%d%d",&n,&m); mem(dp,);

        dp[m][]=; dp[m][]=((p[m]-dp[m][])%MOD+MOD)%MOD;

        FO(i,,m) dp[i][]=p[i];

        FOR(i,m+,n) dp[i][]=(dp[i-][]*+dp[i--m][])%MOD, dp[i][]=((p[i]-dp[i][])%MOD+MOD)%MOD;

        printf("%lld\n",dp[n][]);

    }

    return ;

}

相关文章