ZOJ-3725 Painting Storages DP

　　题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3725

　　n个点排列，给每个点着色，求其中至少有m个红色的点连续的数目。f[i]表示前i个点至少有m个连续红色的个数，则f[i]=f[i-1]*2+2^(i-m-1)-f[i-m-1]。

//STATUS:C++_AC_120MS_1784KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef long long LL;
typedef unsigned long long ULL;
//const
const int N=100010;
const int INF=0x3f3f3f3f;
const int MOD=1000000007,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End

LL f[N],b[N];
int n,m;

int main()
{
 //   freopen("in.txt","r",stdin);
    int i,j,k;
    b[0]=1;
    for(i=1;i<N;i++)b[i]=2*b[i-1]%MOD;
    while(~scanf("%d%d",&n,&m))
    {
        for(i=0;i<m;i++)f[i]=0;
        f[m]=1,f[m+1]=3;
        for(i=m+2;i<=n;i++){
            f[i]=(2*f[i-1]+b[i-m-1]-f[i-m-1])%MOD;
        }
        printf("%lld
",(f[n]+MOD)%MOD);
    }
    return 0;
}