Pop-group “Pink elephant” entered on recording their debut album. In fact they have only two songs: “My love” and “I miss you”, but each of them has a large number of remixes.
The producer of the group said that the album should consist of n remixes. On second thoughts the musicians decided that the album will be of interest only if there are no more than a remixes on “My love” in a row and no more than b remixes on “I miss you” in a row. Otherwise, there is a risk that even the most devoted fans won’t listen to the disk up to the end.
How many different variants to record the album of interest from n remixes exist? A variant is a sequence of integers 1 and 2, where ones denote remixes on “My love” and twos denote remixes on “I miss you”. Two variants are considered different if for some i in one variant at i-th place stands one and in another variant at the same place stands two.
Input
The only line contains integers n, a, b (1 ≤ a, b ≤ 300; max( a, b) + 1 ≤ n ≤ 50 000).
Output
Output the number of different record variants modulo 10 9+7.
Example
| input | output |
|---|---|
3 2 1 |
4 |
Notes
In the example there are the following record variants: 112, 121, 211, 212.
题意
一共有n个1 2，连续1 2 分别不能超过a b 个 问有几种写法
思路
DP
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <cmath>
using namespace std;
long long dp[5][50005] = {0};
int n, m, i, j, k, a1, b1;int main()
{scanf("%d %d %d", &n, &a1, &b1);dp[1][0] = dp[2][0] = 1;for(i = 1; i <= n; i++){for(j = max(i - a1, 0); j < i; j++){dp[1][i] = (dp[1][i] + dp[2][j]) % 1000000007;}for(j = max(i - b1, 0); j < i; j++){dp[2][i] = (dp[2][i] + dp[1][j]) % 1000000007;}}long long ans = (dp[1][n] + dp[2][n]) % 1000000007;printf("%lld\n", ans);return 0;
}