ZOJ 3725 Painting Storages（很好的dp题）

There is a straight highway with N storages alongside it labeled by 1,2,3,...,N. Bob asks you to paint all storages with two colors: red and blue. Each storage will be painted with exactly one color.

Bob has a requirement: there are at least M continuous storages (e.g. "2,3,4" are 3 continuous storages) to be painted with red. How many ways can you paint all storages under Bob's requirement?

Input

There are multiple test cases.

Each test case consists a single line with two integers: N and M (0<N, M<=100,000).

Process to the end of input.

Output

One line for each case. Output the number of ways module 1000000007.

Sample Input

4 3 

Sample Output

3

题目链接：Click here~~

题意：

n个格子排成一条直线，可以选择涂成红色或蓝色，问至少有m 个连续格子为红色的方案数。

解释一下这个：

dp[i]=(dp[i-1]*2%mod+mod_pow(2,i-m-1,mod)-dp[i-m-1]%mod+mod)%mod;

dp[i-1]*2指前i-1个已经符合条件，第i个可红可蓝，所以*2

mod_pow(2,i-m-1,mod)-dp[i-m-1]%mod指的是前i-1个还不符合条件，但是加上第i位（红色）就符合了条件的情况数。第i-m位必须为蓝（隔断），前i-m-1位可红可蓝，所以mod_pow(2,i-m-1,mod)，但是要排除前i-m-1位符合条件的情况，所以-dp[i-m-1]。第i-m位的后面都是红色格子。

#include<iostream>  #include<algorithm>  #include<string>  #include<map>  #include<string.h>  #include<vector>  #include<cmath>  #include<stdlib.h>  #include<cstdio>  #define ll long long  using namespace std;long long dp[100010];long long mod=1000000007;long long mod_pow(long long x,long long n,long long mod){long long res = 1;while(n > 0){if(n&1)res = res * x %mod;x = x * x % mod;n >>= 1;}return res;}int main(){long long n,m;while(scanf("%lld%lld",&n,&m)!=EOF){memset(dp,0,sizeof(dp));dp[m]=1;  //当长度刚好只有m的时候，当然只有1种排法for(int i=m+1;i<=n;i++){dp[i]=(dp[i-1]*2%mod+mod_pow(2,i-m-1,mod)-dp[i-m-1]%mod+mod)%mod;}cout<<dp[n]%mod<<endl;}return 0;}