2019牛客暑期多校训练营（第一场）E题 ABBA（DP）

链接：https://ac.nowcoder.com/acm/contest/881/E

题目描述

Bobo has a string of length 2(n + m) which consists of characters `A` and `B`. The string also has a fascinating property: it can be decomposed into (n + m) subsequences of length 2, and among the (n + m) subsequences n of them are `AB` while other m of them are `BA`.

Given n and m, find the number of possible strings modulo $(109+7)(109+7).$

输入描述:

The input consists of several test cases and is terminated by end-of-file.

Each test case contains two integers n and m.

* $\mathrm{0\le n,m\le 1030\le n,m\le 103*\; There\; are\; at\; most\; 2019\; test\; cases,\; and\; at\; most\; 20\; of\; them\; has\; max\{n,m\}>50max\{n,m\}>50.}$

输出描述:

For each test case, print an integer which denotes the result.

示例1

输入

复制

1 2
1000 1000
0 0

输出

复制

13
436240410
1
思路：一开始讨论，想的是个找规律（虽然我读不懂题，也猜不对）。后来看了题解，才知道是个dp。
          是酱紫的，题目说有N个AB，M个BA，那你就要注意到（A的个数）=（B的个数），尝试在A或B后面加A、B，但是这样很麻烦，所以就直接在A后面加AB，在B后面加BA。其中：DP的时候（A的个数  ）<=（n+B的个数），（B的个数  ）<=（m+A的个数）。首先你要选取A在开头或是B在开头，然后选他们后面跟了什么（A后面尝试加AB，试试有多少种可能）（B后面尝试加BA，试试有多少种可能），然后把两者相加，就是答案。
*我队友：“来自那啥的肯定”

#include <cstdio>
#include <iostream>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
#include <map>
using namespace std;
#define ll long long
int mod = 1e9+7;

int dp[5000+8][5000+8];

int main()
{
    int n, m;
    while(~scanf("%d%d", &n, &m))
    {
        for(int i = 0; i <= n+m; i++)
            for(int j = 0; j <= n+m; j++)
                dp[i][j] = 0;
        dp[0][0] = 1;
        for(int i = 0; i <= n+m; i++)//A的个数
        {
            for(int j = 0; j <= m+n; j++)//B的个数
            {
                int cA = i;//尝试插入A
                int needAB = n-cA;//A后面连接多少个AB
                if(needAB>n+m-j)continue;
                else
                    dp[i+1][j] = (dp[i+1][j]+dp[i][j])%mod;
                int cB = j;//尝试插入B
                int needBA = m-cB;//B后面连接多少个BA
                if(needBA>n+m-i)continue;
                else
                    dp[i][j+1] = (dp[i][j+1]+dp[i][j])%mod;
            }
        }
        printf("%d
", dp[n+m][n+m]);
    }
    return 0;
}