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]。

 1 //STATUS:C++_AC_120MS_1784KB

 2 #include <functional>

 3 #include <algorithm>

 4 #include <iostream>

 5 //#include <ext/rope>

 6 #include <fstream>

 7 #include <sstream>

 8 #include <iomanip>

 9 #include <numeric>

10 #include <cstring>

11 #include <cassert>

12 #include <cstdio>

13 #include <string>

14 #include <vector>

15 #include <bitset>

16 #include <queue>

17 #include <stack>

18 #include <cmath>

19 #include <ctime>

20 #include <list>

21 #include <set>

22 #include <map>

23 using namespace std;

24 //using namespace __gnu_cxx;

25 //define

26 #define pii pair<int,int>

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

28 #define lson l,mid,rt<<1

29 #define rson mid+1,r,rt<<1|1

30 #define PI acos(-1.0)

31 //typedef

32 typedef long long LL;

33 typedef unsigned long long ULL;

34 //const

35 const int N=100010;

36 const int INF=0x3f3f3f3f;

37 const int MOD=1000000007,STA=8000010;

38 const LL LNF=1LL<<60;

39 const double EPS=1e-8;

40 const double OO=1e15;

41 const int dx[4]={-1,0,1,0};

42 const int dy[4]={0,1,0,-1};

43 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};

44 //Daily Use ...

45 inline int sign(double x){return (x>EPS)-(x<-EPS);}

46 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}

47 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}

48 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}

49 template<class T> inline T Min(T a,T b){return a<b?a:b;}

50 template<class T> inline T Max(T a,T b){return a>b?a:b;}

51 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}

52 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}

53 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}

54 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}

55 //End

56 

57 LL f[N],b[N];

58 int n,m;

59 

60 int main()

61 {

62  //   freopen("in.txt","r",stdin);

63     int i,j,k;

64     b[0]=1;

65     for(i=1;i<N;i++)b[i]=2*b[i-1]%MOD;

66     while(~scanf("%d%d",&n,&m))

67     {

68         for(i=0;i<m;i++)f[i]=0;

69         f[m]=1,f[m+1]=3;

70         for(i=m+2;i<=n;i++){

71             f[i]=(2*f[i-1]+b[i-m-1]-f[i-m-1])%MOD;

72         }

73         printf("%lld\n",(f[n]+MOD)%MOD);

74     }

75     return 0;

76 }