2019 cattle off more school first field E ABBA (DP) solution to a problem

The meaning of problems: ask you a length of 2 * (n + m) by a string of (n + m) th A and B, AB required sequence of n and m BA sequence, there are several such strings

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

Ideas:

Suppose there is a valid string, because the sub-sequence of n and m AB BA, then obviously there will be for the first n A of AB A, B must first m of the BA B. Because if I was the first n A there is a BA of A, I was not able to easily find a more back from A to B with this, then obviously the first n A must for the AB of A.

We assume DP [i] [j] for the i-th Aj B, let A:

If i <n that it can be placed directly A, above reason

If i> = n, then we need to ensure that when A and B give BA with the front, then the current needed BA A is min (j, m) number, he has given the i - n a, so that (i - n ) <min (j, m) can then continue to discharge.

Code:

#include<cmath>
#include<set>
#include<map>
#include<queue>
#include<cstdio>
#include<vector>
#include<cstring>
#include <iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 2e3 + 5;
const int M = 50 + 5;
const ull seed = 131;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
ll dp[maxn][maxn];
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++){
            for(int j = 0; j <= n + m; j++){
                if(i < n || (i - n) < j){   //push A
                    dp[i + 1][j] = (dp[i + 1][j] + dp[i][j]) % MOD;
                }
                if(j < m || (j - m) < i){   //push B
                    dp[i][j + 1] = (dp[i][j + 1] + dp[i][j]) % MOD;
                }
            }
        }
        printf("%lld\n", dp[n + m][n + m]);
    }
    return 0;
}