Problem:
Score : 100 points

Problem Statement

There are N children, numbered 1, 2, \ldots, N.

They have decided to share K candies among themselves.
Here, for each i (1 \leq i \leq N), Child i must receive between 0 and a_i candies (inclusive).
Also, no candies should be left over.

Find the number of ways for them to share candies, modulo 10^9 + 7.
Here, two ways are said to be different when there exists a child who receives a different number of candies.

Constraints

All values in input are integers.

1 \leq N \leq 100

0 \leq K \leq 10^5

0 \leq a_i \leq K

Input

Input is given from Standard Input in the following format:

N K
a_1 a_2 \ldots a_N

Output

Print the number of ways for the children to share candies, modulo 10^9 + 7.

Sample Input 1

3 4
1 2 3

Sample Output 1

5

There are five ways for the children to share candies, as follows:

(0, 1, 3)

(0, 2, 2)

(1, 0, 3)

(1, 1, 2)

(1, 2, 1)

Here, in each sequence, the i-th element represents the number of candies that Child i receives.

Sample Input 2

1 10
9

Sample Output 2

0

There may be no ways for the children to share candies.

Sample Input 3

2 0
0 0

Sample Output 3

1

There is one way for the children to share candies, as follows:

(0, 0)

Sample Input 4

4 100000
100000 100000 100000 100000

Sample Output 4

665683269

Be sure to print the answer modulo 10^9 + 7.

Solution in C (Clang 3.8.0) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
 
long int N,K,A[101][100001],a[101];

long int dp(long int n,long int m)
{
  long int k,temp,s;
  
  if(A[n][m]!=-1)
  {
    return A[n][m];
  }
  else
  {
    if(a[n]!=0)
    {
      k=m/a[n];
      temp=0;
      for(s=0;s<k+1;s++)
      {
        temp+=dp(n-1,m-s*a[n]);
      }   
    }
    else
    {
      temp=dp(n-1,m);
    }
  }
  
  temp=temp%1000000007;
  A[n][m]=temp;
  
  return temp;
  
}
 
int main()
{
  
  long int i,j,x;
  long int total;
  scanf("%ld %ld",&N,&K);
  

  for(i=1;i<N+1;i++)
  {
    scanf("%ld",&a[i]);
  }
  
  for(i=0;i<N+1;i++)
  {
    for(j=0;j<K+1;j++)
    {
      A[i][j]=-1;
    }
  }
  
  x=0;
  for(i=1;i<N+1 && x==0;i++)
  {
    if(a[i]!=0)
    {
      x=1;
      for(j=0;j<K+1;j++)
      {
        if(j%a[i]==0)
        {
          A[i][j]=1;
        }
        else
        {
          A[i][j]=0;
        }
      }
    }
  }
  
  if(x==0)
  {
    printf("0");
  }
  else
  {
    printf("%ld",dp(N,K));
  }
  
  return 0;
}