CodeForces 1288C（DP）

Description：

You are given two integers $n$ and $m$ . Calculate the number of pairs of arrays $(a,b)$ such that:

the length of both arrays is equal to $m$ ;
each element of each array is an integer between $1$ and $n$ (inclusive);
ai≤bi for any index i from $1$ to $m$ ;
array $a$ is sorted in non-descending order;
array $b$ is sorted in non-ascending order.
As the result can be very large, you should print it modulo $10^9+7$ .

Input

The only line contains two integers $n$ and $m$ ( $1≤n≤1000, 1≤m≤10$ ).

Output

Print one integer – the number of arrays $a$ and $b$ satisfying the conditions described above modulo $10^9+7$ .

Examples

inputCopy

2 2

outputCopy

inputCopy

10 1

outputCopy

inputCopy

723 9

outputCopy

157557417

Note

In the first test there are 5 suitable arrays:

$a=[1,1],b=[2,2];$
$a=[1,2],b=[2,2];$
$a=[2,2],b=[2,2];$
$a=[1,1],b=[2,1];$
$a=[1,1],b=[1,1].$

题意：

让你构造出两个序列，这两个序列中的元素都不能大于 $n$ 长度为 $m$ ，并且 $a$ 序列的元素相对位置要小于等于 $b$ 。求一共有多少种情况。
直接二维dp， $dp[i][j]$ 表示序列第i位是j的种类数。

AC代码：

#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <queue>
using namespace std;
#define sd(n) scanf("%d", &n)
#define sdd(n, m) scanf("%d%d", &n, &m)
#define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define pd(n) printf("%d\n", n)
#define pc(n) printf("%c", n)
#define pdd(n, m) printf("%d %d", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n, m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld", &n)
#define sldd(n, m) scanf("%lld%lld", &n, &m)
#define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)
#define sf(n) scanf("%lf", &n)
#define sc(n) scanf("%c", &n)
#define sff(n, m) scanf("%lf%lf", &n, &m)
#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define ss(str) scanf("%s", str)
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define mem(a, n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mod(x) ((x) % MOD)
#define gcd(a, b) __gcd(a, b)
#define lowbit(x) (x & -x)
typedef pair<int, int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
inline int read()
{
	int ret = 0, sgn = 1;
	char ch = getchar();
	while (ch < '0' || ch > '9')
	{
		if (ch == '-')
			sgn = -1;
		ch = getchar();
	}
	while (ch >= '0' && ch <= '9')
	{
		ret = ret * 10 + ch - '0';
		ch = getchar();
	}
	return ret * sgn;
}
inline void Out(int a) //Êä³öÍâ¹Ò
{
	if (a > 9)
		Out(a / 10);
	putchar(a % 10 + '0');
}

ll gcd(ll a, ll b)
{
	return b == 0 ? a : gcd(b, a % b);
}

ll lcm(ll a, ll b)
{
	return a * b / gcd(a, b);
}
///快速幂m^k%mod
ll qpow(ll a, ll b, ll mod)
{
	if (a >= mod)
		a = a % mod + mod;
	ll ans = 1;
	while (b)
	{
		if (b & 1)
		{
			ans = ans * a;
			if (ans >= mod)
				ans = ans % mod + mod;
		}
		a *= a;
		if (a >= mod)
			a = a % mod + mod;
		b >>= 1;
	}
	return ans;
}

// 快速幂求逆元
int Fermat(int a, int p) //费马求a关于b的逆元
{
	return qpow(a, p - 2, p);
}

///扩展欧几里得
int exgcd(int a, int b, int &x, int &y)
{
	if (b == 0)
	{
		x = 1;
		y = 0;
		return a;
	}
	int g = exgcd(b, a % b, x, y);
	int t = x;
	x = y;
	y = t - a / b * y;
	return g;
}

///使用ecgcd求a的逆元x
int mod_reverse(int a, int p)
{
	int d, x, y;
	d = exgcd(a, p, x, y);
	if (d == 1)
		return (x % p + p) % p;
	else
		return -1;
}

///中国剩余定理模板0
ll china(int a[], int b[], int n) //a[]为除数，b[]为余数
{
	int M = 1, y, x = 0;
	for (int i = 0; i < n; ++i) //算出它们累乘的结果
		M *= a[i];
	for (int i = 0; i < n; ++i)
	{
		int w = M / a[i];
		int tx = 0;
		int t = exgcd(w, a[i], tx, y); //计算逆元
		x = (x + w * (b[i] / t) * x) % M;
	}
	return (x + M) % M;
}

ll dp[1010][1010], dp2[1010][1010];
ll ans = 0;
int n, m;
int main()
{
	sdd(n, m);
	mem(dp, 0);
	mem(dp2, 0);
	rep(i, 1, n)
		dp[1][i] = dp2[1][i] = 1;
	rep(i, 2, m)
	{
		rep(j, 1, n)
		{
			rep(k, 1, j)
			{
				dp[i][j] += dp[i - 1][k];
				dp[i][j] %= MOD;
			}
		}
	}
	rep(i, 2, m)
	{
		rep(j, 1, n)
		{
			rep(k, j, n)
			{
				dp2[i][j] += dp2[i - 1][k];
				dp2[i][j] %= MOD;
			}
		}
	}
	rep(i, 1, n)
	{
		rep(j, 1, i)
		{
			ans += dp2[m][i] * dp[m][j];
			ans %= MOD;
		}
	}
	pld(ans);
	return 0;
}