DP 2019牛客暑期多校训练营 E.ABBA

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本文介绍了一种计算特定条件下字符串组合数量的方法，通过定义状态和转移方程进行动态规划求解，适用于长度为2*(n+m)且由字符A和B组成的字符串，其中n个子序列不含AB，m个子序列不含BA。

E.ABBA

题意

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

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.

* 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.

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

分析

字符串成立的条件

贪心

A肯定先用来给AB，再用来做BA

即前n个A做AB，后m个A做BA

对于第n+i个A，前面至少存在i个B

Pa-n<=Pb

B肯定先用来给BA，再用来给AB

即前m个B做BA，后n个B做AB

对于第m+i个B，前面至少存在i个A

Pb-m<=Pb

DP

状态

状态很好定义dp[x+y][x],dp[x][y]都可以

转移方程

if (x + 1 - n <= y) dp[x + 1][y] += dp[x][y]; //符合条件，则转移

if (y + 1 - m <= x) dp[x][y + 1] += dp[x][y]; //符合条件，则转移

代码

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 1010;
const int mod = 1e9 + 7;
int dp[maxn << 1][maxn << 1];
int main() {
	int n, m, sum, limit;
	while (~scanf("%d%d", &n, &m)) {
		sum = n + m;
		dp[0][0] = 1;    //初始化
		for (int x = 0; x <= sum; x++) {
			for (int y = 0; y <= sum; y++) {
				if (x + 1 - n <= y)    //转移方程
					dp[x + 1][y] += dp[x][y], dp[x + 1][y] %= mod;
				if (y + 1 - m <= x)
					dp[x][y + 1] += dp[x][y], dp[x][y + 1] %= mod;
			}
		}
		printf("%d\n", dp[n + m][n + m]);
		for (int i = 0; i <= sum + 1; i++)    //初始化
			for (int j = 0; j <= sum + 1; j++)
				dp[i][j] = 0;
	}
}