ZOJ - 3725 Painting Storages

最新推荐文章于 2019-05-06 14:06:31 发布

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最新推荐文章于 2019-05-06 14:06:31 发布

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原文链接：http://www.cnblogs.com/gavanwanggw/p/6741900.html

本文探讨了一个有趣的问题：如何计算在一条直线上排列的N个存储单元中，将它们涂成红蓝两色，使得至少有M个连续的红色存储单元的方案数量。采用动态规划的方法进行解答，并提供了完整的C++代码实现。

Description

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&ltN, 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

题意：n个格子排成一条直线，能够选择涂成红色或蓝色，问最少 m 个连续为红色的方案数。

思路：DP，分两种情况，一种是对于第i个，假设前i-1个已经有了，那么第i个就无所谓了。还有一种是加上第i个才干构成m个连续的话，那么第i-m个就是蓝色的，然后让前i-1-m个不包括连续m个的红色。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
typedef long long ll;
using namespace std;
const int maxn = 100005;
const int mod = 1000000007;

ll f[maxn], dp[maxn];
int n, m;

int main() {
	f[0] = 1;
	for (int i = 1; i < maxn; i++)		
		f[i] = f[i-1] * 2 % mod;

	while (scanf("%d%d", &n, &m) != EOF) {
		if (m > n) {
			printf("0\n");
			continue;
		}

		memset(dp, 0, sizeof(dp));
		dp[m] = 1;
		for (int i = m+1; i <= n; i++) 
			dp[i] = ((dp[i-1] * 2 + f[i-1-m] - dp[i-m-1]) % mod + mod) % mod;

		printf("%lld\n", dp[n]);
	}
	return 0;
}

转载于:https://www.cnblogs.com/gavanwanggw/p/6741900.html