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#include <bits/stdc++.h>
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using namespace std;
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/*
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http://oj.521ma.cn/problem.php?id=1302
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视频讲解
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https://www.bilibili.com/video/av88194897/
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感想：原来只要掌握基本背包问题，一些变换的题型也可以解决。
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状态转移方程：
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i:i件物品
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j:在j这个大的空间内
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dp 状态表示 f(i,j) --->集合---->从前i个数中选且和为m的所有选法
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--->属性---->方案数，即数量 count
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dp 状态计算 第i个数选还是不选 ----> 不选-->f(i-1,j)
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----> 选 -->f(i-1,j-a[i])
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把选与不选的总个数加在一起，就是最终想要的结果
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f(i,j)=f(i-1,j)+f(i-1,j-ai) 二维转一维---->f(j)=f(j)+f(j-ai)
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*/
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const int N = 1001;
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int a[N], f[N];
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int n, m;
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int main() {
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//输入+输出重定向
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freopen("../1302.txt", "r", stdin);
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cin >> n >> m;
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for (int i = 1; i <= n; i++) cin >> a[i];
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//初始化 因为如果和为0，那也就是什么都不放，那就是1种方案，所以初始化f[0]=1,
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f[0] = 1;
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//遍历每一个数字
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for (int i = 1; i <= n; i++)
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for (int j = m; j >= a[i]; j--) //和01背包一样，需要倒序
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f[j] = f[j] + f[j - a[i]]; //求count问题
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cout << f[m] << endl; //输出结果
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//关闭文件
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fclose(stdin);
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return 0;
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}
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