题意

给你一个数字$nn$，只是它的表示形式是它的素数，问你$n-1n-1$的素数表示

思路

质因数分解

代码

/**
 *    author: andif
 *    created: 30.07.2023 13:33:18
**/
#include<bits/stdc++.h>
using namespace std;

#define de(x) cerr << #x << " = " << x << endl
#define dd(x) cerr << #x << " = " << x << " "
#define rep(i, a, b) for(int i = a; i < b; ++i)
#define per(i, a, b) for(int i = a; i > b; --i)
#define mt(a, b) memset(a, b, sizeof(a))
#define sz(a) (int)a.size()
#define fi first
#define se second
#define qc ios_base::sync_with_stdio(0);cin.tie(0)
#define eb emplace_back
#define all(a) a.begin(), a.end()
using ll = long long;
using db = double;
using pii = pair<int, int>;
using pdd = pair<db, db>;
using pll = pair<ll, ll>;
using vi = vector<int>;
const db eps = 1e-9;

int main() {
    qc;
    int t; cin >> t;
    while(t--) {
        int k; cin >> k;
        int n = 1;
        rep(i, 0, k) {
            int p, e; cin >> p >> e;
            n *= pow(p, e);
            // dd(p), de(e);
        }
        n--;
        // de(n);
        vi p, e;
        for (int i = 2; i * i <= n; ++i) {
            if (n % i) continue;
            p.eb(i);
            int c = 0;
            while(n % i == 0) {
                c++;
                n /= i;
            }
            e.eb(c);
        }
        if (n != 1) {
            p.eb(n);
            e.eb(1);
        }
        per(i, sz(p) - 1, -1) {
            cout << p[i] << " " << e[i];
            if (!i) cout << '\n';
            else cout << ' ';
        }
    }
    return 0;
}