# Wet Shark and Two Subsequences

### Problem Statement :

```One day, Wet Shark was given an array X = {x1, x2, ..., xm}. As always, he started playing with its subsequences.

When you came to know about this habit, you presented him a task of finding all pairs of subsequences, (A,B), which satisfies all of the following constraints. We will represent a pair of subsequence as A = {xa1,xa2,...,xan} and B = {xb1,xb2,...,xbn}

A and B must be of same length, i.e., |A| = |B|.
Σ(xai + xbi) = r
Σ(xai - xbi) = s
Please help Wet Shark determine how many possible subsequences A and B can exist. Because the number of choices may be big, output your answer modulo 10^9 + 7 =100000007.

Note:

Two segments are different if there's exists at least one index i such that element xi is present in exactly one of them.
Both subsequences can overlap each other.
Subsequences do not necessarily have to be distinct
Input Format

The first line consists of 3 space-separated integers m, r, s, where m denotes the length of the original array, X, and r and s are as defined above.
The next line contains m space-separated integers,  x1, x2,..., xm, representing the elements of X.

Constraints
1 <= m <= 100
0 <= r,s <=2000
1 <= xi <= 2000

Output Format

Output total number of pairs of subsequences, (A,B), satisfying the above conditions. As the number can be large, output it's modulo 10^9 + 7 =100000007.```

### Solution :

```                            ```Solution in C :

In C++ :

/*
*/

#include <fstream>
#include <iostream>
#include <string>
#include <complex>
#include <math.h>
#include <set>
#include <vector>
#include <map>
#include <queue>
#include <stdio.h>
#include <stack>
#include <algorithm>
#include <list>
#include <ctime>
#include <memory.h>
#include <ctime>

#define y0 sdkfaslhagaklsldk
#define y1 aasdfasdfasdf
#define j1 assdgsdgasghsf
#define tm sdfjahlfasfh
#define lr asgasgash

#define eps 1e-9
//#define M_PI 3.141592653589793
#define bs 1000000007
#define bsize 256
#define MAG 10000

using namespace std;

long n,r,s,na,nb,q,dp;
long long ans;

{
a+=b;
if (a>=bs)
a-=bs;
}

int main(){
//freopen("evacuation.in","r",stdin);
//freopen("evacuation.out","w",stdout);
//freopen("input.txt","r",stdin);
//freopen("output.txt","w",stdout);
ios_base::sync_with_stdio(0);
//cin.tie(0);

cin>>n>>r>>s;
nb=(r-s)/2;

na=r-nb;
if (na>2000) return 1;

dp=1;

for (int i=1;i<=n;i++)
{
cin>>q;
for (int j=0;j<=2000;j++)
for (int l=0;l<=n;l++)
{
if (j>=q&&l>0)
}
}
long long answ=0;

for (int p=1;p<=n;p++)
{
ans=dp[n][na][p];
if (s%2!=r%2||r<s)
ans=0;else
ans*=dp[n][nb][p];
ans%=bs;
answ=answ+ans;
answ%=bs;}
cout<<answ<<endl;

cin.get();cin.get();
return 0;}

In Java :

import java.io.*;
import java.util.*;

public class Solution {
private static PrintWriter out;
public static int mod = 1000000007;

public static void main(String[] args) throws IOException {
out = new PrintWriter(System.out, true);

int M = in.nextInt();
int R = in.nextInt();
int S = in.nextInt();
if ((R+S) % 2 != 0 || S >= R) {
out.println("0");
out.close();
System.exit(0);
}

int P = (R+S)/2, Q = (R-S)/2;

long[][] nways = new long[M+1][P+1];
nways = 1;
for (int i = 1; i <= M; i++) {
int x = in.nextInt();
for (int j = M; j >= 1; j--) {
for (int k = P; k >= x; k--) {
nways[j][k] += nways[j-1][k-x];
if (nways[j][k] >= mod) nways[j][k] -= mod;
}
}
}

long total = 0;
for (int i = 0; i <= M; i++) {
total = (total + nways[i][P] * nways[i][Q]) % mod;
}
out.println(total);
out.close();
System.exit(0);

}

public StringTokenizer tokenizer;

tokenizer = null;
}

public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
} catch (IOException e) {
throw new RuntimeException(e);
}
}
}

public int nextInt() {
return Integer.parseInt(next());
}
}

}

In C :

#include <stdio.h>
#include <stdlib.h>
#define MOD 1000000007
int a;
long long dp1;

int main(){
int m,r,s,A,B,i,j,k;
long long ans=0;
scanf("%d%d%d",&m,&r,&s);
for(i=0;i<m;i++)
scanf("%d",a+i);
if((r+s)%2 || r<s){
printf("0");
return 0;
}
A=(r+s)/2;
B=(r-s)/2;
for(i=0;i<=m;i++)
for(j=0;j<m;j++)
for(k=0;k<=A;k++)
if(!i)
if(!k)
dp1[i][j][k]=1;
else
dp1[i][j][k]=0;
else if(!j){
if(i>1)
dp1[i][j][k]=0;
else if(k==a[j])
dp1[i][j][k]=1;
else
dp1[i][j][k]=0;
}
else if(i>j+1)
dp1[i][j][k]=0;
else{
dp1[i][j][k]=dp1[i][j-1][k];
if(k-a[j]>=0)
dp1[i][j][k]=(dp1[i][j][k]+dp1[i-1][j-1][k-a[j]])%MOD;
}
for(i=1;i<=m;i++)
ans=(ans+dp1[i][m-1][A]*dp1[i][m-1][B]%MOD)%MOD;
printf("%lld",ans);
return 0;
}

In Python3 :

def solve():
mod = 10**9+7
m,r,s = map(int,input().rstrip().split(' '))
if m == 0 or r<=s: return 0
arr = list(map(int,input().rstrip().split(' ')))
if (r-s) % 2 != 0: return 0
sumb = (r-s)//2
suma = r - sumb

def get_ways(num):
dp = [*(num+1) for _ in range(m+1)]
dp = 1
for c in arr:
for i in range(len(dp)-1,c-1,-1):
for j in range(len(dp)-2,-1,-1):
if dp[j][i-c]>0:
#print(j,i,c)
dp[j+1][i]+=dp[j][i-c]
#print(dp)
for j in range(len(dp)):
if dp[j][num] > 0:
#print(j,dp[j][num])
yield (j,dp[j][num]) #num coins, count

a = list(get_ways(suma))
#print('---',suma,sumb)
b = list(get_ways(sumb))
res = 0
for i in a:
for j in b:
if i==j: #same length
res += i*j
return res % mod

print(solve())```
```

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