ZOJ - 3725 Painting Storages
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<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: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; }
posted on 2017-04-21 09:26 gavanwanggw 阅读(157) 评论(0) 编辑 收藏 举报