zoj3725s

Painting Storages

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Time Limit: 2 Seconds Memory Limit: 65536 KB

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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

题意：给长度为n的空间，每个空间可以涂成红色和蓝色，问有多少种方案满足有连续m个以上红色空间的方案。

思路：把红色用O蓝色用X表示好了，样例满足的条件就有三种：

OOOO

OOOX

XOOO

递推公式为dp[m]=1;

dp[i]=dp[i-1]*2+2^(i-m-1)-dp[i-m-1] (i>m)

当n刚好等于m的时候是只有一种情况，然后dp[i-1]*2，即是在之前满足的方案后面添O或X，不影响结果，后面是对没满足的情况进行处理，2^(i-m-1)是将i-m的位置涂成X之后i-m-1位置以前的所有满足或不满足情况，接着减去i-m-1之前的满足情况，即使i-m-1的不满足情况，加一个X和m个O，凑成满足情况

#include<stdio.h>
#include<string.h>
#define mod 1000000007
#define LL long long
LL pow[100005],dp[100005];
void init(){
	pow[0]=1;
	for(int i=1;i<=100000;i++)
		pow[i]=pow[i-1]*2%mod;	
}
int main(void){
	init();
	int n,m;
	LL ans;
	while(~scanf("%d",&n)){
		memset(dp,0,sizeof(dp));
		dp[m]=1;
		for(int i=m+1;i<=n;i++)
			dp[i]=((dp[i-1]*2%mod+pow[i-m-1]-dp[i-m-1])%mod+mod)%mod;
		printf("%lld\n",dp[n]);
		
	}	
	return 0;
}