XTU 1233 Coins(DP)

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 ;
}