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 }
