题解,求关注

#include <iostream>
#include <algorithm>
using namespace std ;
int dp[ 35 ][ 550 ];
int main ()
{
    int n, m;
    cin >> n >> m;
    int sum = 0 ;
    dp[ 0 ][ 0 ] = 1 ;
    for ( int i = 1 ; i <= n; i++)
    {
        for ( int j = 0 ; j <= (i+ 1 )*i/ 2 ; j++)
        {
        //dp[i][j] 表⽰ 1~i 的数字拆分成两个⼦数组求和的差的绝对值为 j 的⽅案数。
        dp[i][j] = dp[i -1 ][j+i] + dp[i -1 ][ abs (j-i)];
        }
    }
    if (m == 0 )
        cout << dp[n][m] / 2 << endl ;
    else
        cout << dp[n][m] << endl ;
    return 0 ;
}