思路：前缀和+双指针

代码：

#include <bits/stdc++.h>
using namespace std;
using i64 = int64_t;
int main() {cin.tie(nullptr)->sync_with_stdio(false);cout << fixed << setprecision(20);int t = 1;for (int ti = 0; ti < t; ti += 1) {int n, W;cin >> n >> W;vector<i64> w(n + 1), d(n + 1);for (int i = 1; i <= n; i += 1) {cin >> w[i];w[i] += w[i - 1];}for (int i = 1; i <= n; i += 1) {cin >> d[i];d[i] += d[i - 1];}i64 ans = -1e18, mn = 1e18;for (int i = 1, j = 0; i <= n; i += 1) {while (w[i] - w[j] >= W) {mn = min(mn, d[j]);j += 1;}ans = max(ans, d[i] - mn);}cout << ans;}
}

E. Increasing Subsequences

思路：首先如果有0,1,2……,x-1,那么有2^x种选择方案.考虑二进制拆分,首先构造出最高位的答案2^x种.然后从大到小遍历每一位,如果第i位是1,就在序列最后添加一个i,不难发现每添加一次数会让答案增加2^i,而且因为添加的数是递减的,所以不会相互干扰.

代码：

void solve(){int n;cin >> n;int t = __lg(n);vector<int> ans;for(int i = 0; i < t; i++){ans.push_back(i);}for(int i = t - 1; i >= 0; i--){if (n >> i & 1){ans.push_back(i);}}cout << ans.size() << '\n';for(auto x : ans) cout << x << ' ';cout << '\n';
}