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-7
# define MOD 1000000007
# define INF 1000000000
# 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<<1,l,mid
# define rch p<<1|1,mid+1,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=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
void Out(int a) {
    if(a<0) {putchar(‘-‘); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+‘0‘);
}
const int N=1000005;
//Code begin...

LL dp[N][2], p[N];

void init(){p[0]=1; FO(i,1,N) p[i]=p[i-1]*2%MOD;}
int main ()
{
    int T, n, m;
    scanf("%d",&T); init();
    while (T--) {
        scanf("%d%d",&n,&m); mem(dp,0);
        dp[m][1]=1; dp[m][0]=((p[m]-dp[m][1])%MOD+MOD)%MOD;
        FO(i,0,m) dp[i][0]=p[i];
        FOR(i,m+1,n) dp[i][1]=(dp[i-1][1]*2+dp[i-1-m][0])%MOD, dp[i][0]=((p[i]-dp[i][1])%MOD+MOD)%MOD;
        printf("%lld\n",dp[n][1]);
    }
    return 0;
}