恢复排列(Restore Permutation) 题解

思路分析

按题目要求使用对应数据结构与算法模拟/递推求解，注意数据范围边界与时间复杂度。

参考代码

#include <iostream>
#include <vector>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    if (!(cin >> n)) return 0;
    vector<long long> s(n + 1);
    for (int i = 1; i <= n; ++i) cin >> s[i];
    vector<long long> bit(n + 1, 0);
    auto add = [&](int idx, long long v) {
        for (int i = idx; i <= n; i += i & -i) bit[i] += v;
    };
    for (int i = 1; i <= n; ++i) add(i, i);
    vector<int> p(n + 1);
    for (int i = n; i >= 1; --i) {
        int pos = 0;
        long long rem = s[i];
        for (int k = 1 << 18; k > 0; k >>= 1) {
            if (pos + k <= n && bit[pos + k] <= rem) {
                pos += k;
                rem -= bit[pos];
            }
        }
        p[i] = pos + 1;
        add(p[i], -p[i]);
    }
    for (int i = 1; i <= n; ++i) {
        cout << p[i] << (i == n ? "\n" : " ");
    }
    return 0;
}