#include #include #include #include #include #include #include #include using namespace std; using namespace atcoder; // using mint = modint998244353; using mint = modint1000000007; using vi = vector; using vvi = vector>; using ll = long long; template using max_heap = priority_queue; template using min_heap = priority_queue, greater<>>; #define rep(i, n) for (int i = 0; i < (int)(n); i++) #define rep2(i, f, n) for (int i = (int) f; i < (int)(n); i++) #define repd(i, n, l) for (int i = (int) n; i >= (int) l; i--) #define all(p) p.begin(),p.end() vector> dydx{{-1, 0}, {1, 0}, {0, -1 }, {0, 1}}; const ll inf = 1LL << 60; ll op(ll a,ll b) {return a+b;} ll e() {return 0LL;} int main() { int n, m; cin >> n >> m; vector A(n+1); vector B(101, 0); rep(i, n){ ll a; cin >> a; a /= 100; A[i+1] = a; } A[0] = 0; sort(all(A)); rep(i, n) A[i+1] += A[i]; rep(i, m){ int b; cin >> b; B[100-b]++; } B[100] = 1000000; // segtree seg(A); for (int i = 1; i <= n; i++){ ll now = i; ll ans = 0; for (int j = 1; j <= 100; j++){ ll d = B[j]; if (d == 0) continue; ll left = max(0LL, now-d); ans += j * (A[now] - A[left]); // cout << ans << ' ' << now << ' ' << left << endl; now = left; if (now == 0) break; } cout << ans << endl; } return 0; }