CodeForces1288 C.Two Arrays(dp/组合数学)

C.Two Arrays
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 109+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 109+7.

Examples
input
2 2

output
5

input
10 1

output
55

input
723 9

output
157557417

题意

题目可以转换成求一个长度为2m的不递减数组，元素取值为1-n；

有两种方法，分别是dp和组合数。

题解一

dp[i][j]，i为长度，j为末尾的取值。

dp[i][j]=sigma{dp[i-1][k](1<=k<=j)}=dp[i][j-1]+dp[i-1][j];

题解二

1-n中，每个数可以取多次，总的取得次数为2m，设i取的次数为xi，

则x1+x2+..xm=2m，求该方程非负整数解的个数，利用隔板法可得ans=C（n-1+2m,2m)

代码一

#include<bits/stdc++.h>
ll dp[21][1001];
int main()
{
  ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
  int n,m;
  cin>>n>>m;
  m*=2;
  dp[1][0]=1;
  rep(i,1,m)
     rep(j,1,n){
         dp[i][j]=dp[i][j-1]+dp[i-1][j];dp[i][j]%=mod;
     }
  ll ans=0;
  rep(i,1,n) ans+=dp[m][i],ans%=mod;
  cout<<ans;
  return 0;
}

代码二

#include<bits/stdc++.h>
using namespace std;
int main()
{
  ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
  int n,m,c[2000][22];
  cin>>n>>m;
  rep(i,0,n-1+2*m) c[i][0]=1;
  rep(i,1,n-1+2*m)
    rep(j,1,min(i,2*m))
      c[i][j]=(c[i-1][j-1]+c[i-1][j])%mod;
  cout<<c[n-1+2*m][2*m];
  return 0;
}