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

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  
#include  
#include  
#include  
#include  
#include  
#include  
#include  
#include  
#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<


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