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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
const int mod=1e9+7;
void add_self(int& a, int b) {
    a += b;
    if(a >= mod) {
        a -= mod;
    }
}
void sub_self(int& a, int b) {
    a -= b;
    if(a < 0) {
        a += mod;
    }
}

int main() {
    int n, k;
    scanf("%d%d", &n, &k);
    vector<int> dp(k + 1);
    // dp[i] - the number of ways such that we used i candies so far
    dp[0] = 1;
    for(int child = 0; child < n; ++child) {
        int up_to;
        scanf("%d", &up_to);
        vector<int> fake(k + 1);
        for(int used = k; used >= 0; --used) {
            int tmp = dp[used];
            int L = used + 1;
            int R = used + min(up_to, k - used);
            if(L <= R) {
                add_self(fake[L], tmp);
                if(R + 1 <= k) {
                    sub_self(fake[R+1], tmp);
                }
            }
            //~ for(int i = L; i <= R; ++i) {
                //~ add_self(dp[i], tmp);
            //~ }
        }
        int prefix_sum = 0;
        for(int i = 0; i <= k; ++i) {
            add_self(prefix_sum, fake[i]);
            add_self(dp[i], prefix_sum);
        }
    }
    printf("%d\n", dp[k]);
}