牛客網多校__ ABBA

鏈接：https://ac.nowcoder.com/acm/contest/881/E
來源：牛客網
 

時間限制：C/C++ 2秒，其他語言4秒
空間限制：C/C++ 524288K，其他語言1048576K
64bit IO Format: %lld

題目描述

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).

輸入描述:

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

Each test case contains two integers n and m.

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

輸出描述:

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

示例1

輸入

複製

1 2
1000 1000
0 0

輸出

複製

13
436240410
1

dp,由於要生成n個AB，m個BA,所以AB的A肯定是先用的前面的A，剩下的A去組成BA，然後可以發現如果

我們用dp[x][y]，其中x爲前綴有x個A，y爲前綴有y個B，則可以通過之後添加A，或添加B，判斷是否會出現前綴裏的x-y>n如果

出現，則爲不合法，因爲這樣會出現大於n的AB，BA也同理

#include <bits/stdc++.h>

using namespace std;

const int maxn = 2e3+9;
typedef long long ll;
const ll mod = 1e9+7;

ll dp[maxn][maxn];

void add(ll &a,ll b){
    a+=b;
    if(a>=mod)a-=mod;
}

int main(){
    int n,m;
    while(~scanf("%d%d",&n,&m)){
        for(int i=0;i<=n+m;i++){
            if(i<=m)dp[0][i]=1;
            else dp[0][i]=0;
        }
        for(int i=1;i<=n+m;i++){
            for(int j=0;j<=n+m;j++){
                dp[i][j]=0;
                if(i-j<=n){
                    add(dp[i][j],dp[i-1][j]);
                }
                if(j-i<=m){
                    add(dp[i][j],dp[i][j-1]);
                }
            }
        }
        printf("%lld\n",dp[n+m][n+m]);
    }
    return 0;
}