题目链接：

http://acm.hust.edu.cn/vjudge/contest/126546#problem/G

Description

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 ? remixes. On second thoughts the
musicians decided that the album will be of interest only if there are no more than ? remixes on “My
love” in a row and no more than ? 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 ? 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 ? in one variant at ?-th place stands one and in another
variant at the same place stands two.

Input

The only line contains integers ?, ?, ? (1 ≤ ?, ? ≤ 300; max(?, ?) + 1 ≤ ? ≤ 50 000).

Output

Output the number of different record variants modulo 109 + 7.

Examples

3 2 1
4

Explanation

In the example there are the following record variants: 112, 121, 211, 212.

题意：

英语弱鸡表示题意真难看懂.
在n个位置填充1 or 2. 要求连续的1的个数不超过a，连续的2的个数不超过b.
求方案数.

题解：

要考虑每一位可能放1或2，容易想到动态规划：
dp[i][1/2][j]：到第i个位置时，放1且有连续j个1的方案(放2且有连续j个2的方案).
转移方程：
j==1: dp[i][1][1] = Σdp[i-1][2][k] ,(k<=b), 即只放一个1的情况等同于前面有任意个连续2的情况.
dp[i][2][1] = Σdp[i-1][1][k] ,(k<=a), 即只放一个2的情况等同于前面有任意个连续1的情况.
j!=1: dp[i][1/2][j] = dp[i-1][1/2][j-1];

上述思路确定后，考虑数组内存的问题，由于n比较大，直接开三维数组估计会MLE.
由于转移具有线性型(i只能由i-1来). 所以可以用滚动数组压缩到二维数组.
注意考虑压缩后的各种问题：
如果后面需要的值在前面被改变了，那么就先需要记录下来.
对j进行枚举时，由于j用到j-1的值，所以要从大往小枚举.

一个简单的滚动数组dp写了一个小时，一直在瞎处理边界.
归根结底还是太弱了，dp还需多加练习.

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <vector>
#define LL long long
#define eps 1e-8
#define maxn 550
#define mod 1000000007
#define inf 0x3f3f3f3f
#define IN freopen("in.txt","r",stdin);
using namespace std;int n, a, b;
int dp[2][350];int main(int argc, char const *argv[])
{//IN;while(scanf("%d %d %d", &n,&a,&b) != EOF){memset(dp, 0, sizeof(dp));dp[0][1] = dp[1][1] = 1;for(int i=2; i<=n; i++) {/*由于要滚动处理，先记录下会改变的值*/int tmpa = dp[0][1];int tmpb = dp[1][1];dp[0][1] = 0;for(int j=1; j<=b; j++)dp[0][1] = (dp[0][1] + dp[1][j]) % mod;dp[1][1] = tmpa;for(int j=2; j<=a; j++)dp[1][1] = (dp[1][1] + dp[0][j]) % mod;/*由于要滚动处理，需要从后往前覆盖*/for(int j=a; j>=3; j--) {dp[0][j] = dp[0][j-1];}if(a >= 2) dp[0][2] = tmpa;for(int j=b; j>=3; j--) {dp[1][j] = dp[1][j-1];}if(b >= 2) dp[1][2] = tmpb;}int ans = 0;for(int i=1; i<=a; i++)ans = (ans + dp[0][i]) % mod;for(int i=1; i<=b; i++)ans = (ans + dp[1][i]) % mod;printf("%d\n", ans);}return 0;
}