EOJ Monthly 2020.7 Sponsored by TuSimple E.因数串 数学/ 构造

vector<ll> ans;
vector<ll> p(16);
vector<ll> e(16);

int main() {
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf("%lld %lld", &p[i], &e[i]);
    }
    ans.pb(1);
    ll tmp = 1;
    for (int i = 0; i < e[0]; i++) {
        tmp *= p[0];
        ans.pb(tmp);
    }
    for (int i = 1; i < n; i++) {
        int len = ans.size();
        for (int j = 0; j < e[i]; j++) {
            int k = ans.size();
            for (int _ = 0; _ < len; _++) {
                ans.pb(ans[k - _ - 1] * p[i]);
            }
        }
    }
    for (auto& k : ans) {
        printf("%lld\n", k);
    }
}