小V和方程

思路：

然后dp即可

#include <bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
#define inf 0x3f3f3f3f
#define ll long long
#define pb push_back
#define PII pair<int,int>
#define X first
#define Y second
#define MP make_pair
#define PI acos(-1.0)
#define eps 1e-8
const int maxn=1e6+10;
const int mod=998244353;
using namespace std;

int n,m,ans;
ll dp[1005][1005];

ll solve(int all,int k)
{
    if(all==0||k==1)
        return 1;
    if(k>all)
        return solve(all,k);
    return (solve(all,k-1)+solve(all-k,k))%mod;
}

int main()
{
    scanf("%d%d",&n,&m);
    int x=1,y=1;
    for(int i=2;i*i<=m;i++)
    {
        int tmp=0;
        while(m%i==0) tmp++,m/=i;
        if(tmp&1) y*=i,tmp--;
        for(int j=0;j<tmp/2;j++) x*=i;
    }
    if(m>1) y*=m;
    dp[0][0]=1;
    for (int i=1;i<=x;i++)
    {
        for (int j=1;j<=i;j++)
        {
            if(i>=j)
            {
                dp[i][j]=(dp[i-1][j-1]+dp[i-j][j])%mod;
            }
            else
            {
                dp[i][j]=dp[i][i];
            }
        }
    }
    ll ans=0;
    for(int i=1;i<=n;i++) ans+=dp[x][i],ans%=mod;
    printf("%lld",ans);
    // system("pause");
    return 0;
}