题意

给出\(n\)个长条，每个长条保证可以表示为\(2^x\)的形式，问你如果一个宽度为\(w\)的盒子最少要多高才能装下这些长条。

思路

贪心。将长条按照长度从大到小排序，对于每一层我们尽量将它装满再装下一层。

可以用\(multiset\)维护每一层剩余的空间。对于当前要放入盒子的长条，在集合中\(lower\_bound\)得到一个合适的位置，如果没有那就新开一层，答案加一；如果有就将该层的空间减少，还有剩余就再次加入到集合中。

代码

#include <algorithm>
#include <iostream>
#include <cstring>
#include <stack>
#include <cstdlib>
#include <map>
#include <queue>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <set>
#include <ctime>
 
#define rep(i, s, t) for (int i = (s); i < (t); i++)
#define wa std::cerr << "----WARN----" << std::endl;
#define wl std::cerr << "********" << std::endl;
#define wr std::cerr << "~~~~~~~~" << std::endl;
#define PII pair<int, int>
#define mp(a, b) make_pair(a, b)
#define fr first
#define sc second
 
typedef long long LL;
 
const int MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const LL LINF = 1LL * 0x3f3f3f3f3f3f3f3f;
 
inline int read();
 
const int N = 100005;
 
int a[N];
 
void solve() {
    int n, w;
    std::cin >> n >> w;
    rep (i, 0, n) {
        std::cin >> a[i];
    }
    std::sort(a, a + n, std::greater<int>());
    std::multiset<int> s;
    int ans = 0;
    rep (i, 0, n) {
        if (s.empty()) {
            if (a[i] != w) s.insert(w - a[i]);
            ans++;
        } else {
            auto it = s.lower_bound(a[i]);
            if (it == s.end()) {
                if (w != a[i]) s.insert(w - a[i]);
                ans++;
                continue;
            } else {
                int t = *it;
                s.erase(it);
                if (t != a[i]) s.insert(t - a[i]);
            }
        }
    }
    std::cout << ans << std::endl;
}
 
inline int read() {
    int s = 0, w = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-') w = -1;
        ch = getchar();
    }
    while (isdigit(ch)) {
        s = s * 10 + ch - '0';
        ch = getchar();
    }
    return s * w;
}
 
signed main() {
    // freopen("in.txt", "r", stdin);
    int T = 1;
    T = read();
    while (T--) solve();
    return 0;
}