注意多组数据分别输出时，一般都是互不干扰的

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 200010;
int n;
LL S;
int a[N];
int get(LL a, LL b, LL c, LL d)//区间函数，找到区间相同的的元素个数，多次调用时考虑函数解决问题）
{
    return min(b, d) - max(a, c) + 1;//四个端点
}
int main()
{
    scanf("%d%lld", &n, &S);//数据大于10w以上时尽量用scanf输入printf输出，减少时间损耗
    LL sum = 0;
    for (int i = 1; i <= n; i ++ )
    {
        scanf("%d", &a[i]);
        sum += a[i];//前缀和
    }
    for (int i = 1; i <= n; i ++ )
        printf("%d ", a[i] - get(1, a[i], S - (sum - a[i]), S - (n - 1)));
      //  cout << (double)clock() / CLOCKS_PER_SEC << "s";
    return 0;
}