no.of pairs of subsets with given difference

#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<string,int> psi;
typedef priority_queue<int> pqmax;
typedef priority_queue<int,vector<int>,greater<int>> pqmin;
typedef unordered_map<int,int> mii;
typedef unordered_map<long long,long long> mll;
typedef unordered_map<string,int> msi;
typedef unordered_map<char,int> mci;
typedef unordered_set<int> si;
typedef unordered_set<long long> sll;
typedef unordered_set<string> ss;
typedef unordered_set<char> sc;
typedef map<int,int> ormii;
typedef map<long long,long long> ormll;
typedef map<string,int> ormsi;
typedef map<char,int> ormci;
typedef set<int> orsi;
typedef set<long long> orsll;
typedef set<string> orss;
typedef set<char> orsc;
typedef vector<int> vi;
typedef vector<string> vs;
typedef vector<char> vc;
typedef vector<ll> vll;
typedef vector<vector<int>> vvi;
typedef vector<vector<string>> vvs;
typedef vector<vector<bool>> vvb;
typedef vector<vector<ll>> vvll;

#define FOR(i, a, b) for (auto i=a; i<=(b); i++)
#define FORd(i,b,a) for (int i =b; i >= a; i--)
#define sortinc(v)  sort(v.begin(),v.end())
#define sortdec(v)  sort(v.rbegin(),v.rend())
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define pob pop_back
#define pf push_front
#define pof pop_front
#define fi first
#define se second
#define ins insert

const int MOD = 1000000007;
int m(ll k) {return k%MOD;}
//type functions here

int solve(vector<int> &A, int diff) {
    int n=A.size();
    int sum=0;
    for(int x:A) sum+=x;
    int B=(sum-diff)/2;
    vvi dp(n+1,vector<int>(B+1,0));
    for(int i=0;i<=B;i++) dp[0][i]=0;
    for(int i=0;i<=n;i++) dp[i][0]=1;

    for(int i=1;i<=n;i++){
    for(int j=1;j<=B;j++)
    {
        if(j>=A[i-1])//current element is not exceeded the sum to acheieve
        dp[i][j]=(dp[i-1][j])+(dp[i-1][j-A[i-1]]);
        else//if exceeded we have to solely depend on n-1 element current sum
        dp[i][j]=dp[i-1][j];


    }
    }
     return dp[n][B];
}


int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int tc=1;
    //cin>>tc;
    FOR(w,1,tc)
    {
        vi A={1,1,2,3};
        int B=1;
       cout<<solve(A,B)<<endl;//expected 3 subset pairs
    }
    return 0;
}