算法

(数论)

注意到 $n \leqslant d(n) = c \leqslant 1e7$，所以直接筛出 $c$ 即可。

C++ 代码

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define srep(i, s, t) for (int i = s; i < (t); ++i)
#define drep(i, n) for (int i = (n) - 1; i >= 0; --i)
#define dsrep(i, t, s) for (int i = (t) - 1; i >= (s); --i)

using std::cin;
using std::cout;
using ll = long long;

const int N = 1e7 + 10;

int pf[N];
ll d[N];
int ans[N];

void init() {
    std::memset(pf, -1, sizeof pf);
    pf[1] = 1;
    for (int i = 2; i * i < N; ++i) {
        if (pf[i] == -1) {
            pf[i] = i;
            for (int j = i * i; j < N; j += i) {
                if (pf[j] == -1) pf[j] = i; 
            }
        }
    }

    d[1] = 1;
    srep(i, 2, N) {
        if (pf[i] == -1) {
            pf[i] = i;
            d[i] = i + 1;
        }
        else {
            int x = i;
            d[i] = 1;
            while (x % pf[i] == 0) {
                x /= pf[i];
                d[i] = d[i] * pf[i] + 1; 
            }
            d[i] *= d[x];
        }
    }

    std::memset(ans, -1, sizeof ans);
    dsrep(i, N, 1) {
        if (d[i] < N) ans[d[i]] = i;
    }   
}

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);

    init();

    int t;
    cin >> t;

    while (t--) {
        int c;
        cin >> c;
        cout << ans[c] << '\n';
    }

    return 0;
}