Gym - 101911D（转化为求因子）

Quite often the jury of Saratov SU use the problem “Masquerade” in different practice sessions before the contest. This problem is quite easy — all you need is to print the product of two integers which were read from the input stream.

As usual, the jury had prepared this problem once again. The jury had nn testcases, the ii-th testcase was a pair of positive integers aiai and bibi, both integers didn’t exceed 107107. All testcases were pairwise distinct.

Unfortunately, something went wrong. Due to hardware issues all testcases have disappeared. All that the jury were able to restore are the number of testcases nn and the answers to these testcases, i. e. a sequence of nn numbers c1,c2,…,cnc1,c2,…,cn, such that ai⋅bi=ciai⋅bi=ci.

The jury ask you to help them. Can you provide any possible testset? Remember that all testcases were distinct and all numbers in each testcase were positive integers and didn’t exceed 107107.

Input
First line contains one insteger nn (1≤n≤2⋅1051≤n≤2⋅105) — the number of lost testcases.

Second line contains nn space-separated integers c1,c2,…,cnc1,c2,…,cn (1≤ci≤1071≤ci≤107) — the answers to the testcases.

Output
If there is no such testset, print NO.

Otherwise, print YES in first line. Then print nn more lines, the ii-th of them should contain two space separated positive integers aiai and bibi not exceeding 107107. All pairs (ai,bi)(ai,bi) must be distinct, and, for each i∈[1,n]i∈[1,n], the condition ai⋅bi=ciai⋅bi=ci must be met.

Examples
Input
4
1 3 3 7
Output
YES
1 1
1 3
3 1
1 7
Input
5
3 1 3 3 7
Output
NO
Input
6
9 10 9 10 9 10
Output
YES
1 9
1 10
3 3
5 2
9 1
2 5
Note
In the first example one of the possible testsets is (a1=1a1=1, b1=1b1=1), (a2=1a2=1, b2=3b2=3), (a3=3a3=3, b3=1b3=1), (a4=1a4=1, b4=7b4=7).

In the second example a testset consisting of distinct tests doesn’t exist.

大题就是，给你n个数，每个数c对应输出一对数a,b，a*b=c;这些对数不能重复（（1，3）和（3，1）不算）
根据一个数的因子都可以由有限个质因子相乘得来，注意的是c存在大于sqrt©的质因子最多只有一个。所以只需要标记sqrt(1e7）+1内的质数就够用了
一开始不会处理一个数是第几个数（按输入的顺序），因为直接用数组的话要1e7很大了，后来学了vector<pair<int, int> > arr(n), ans(n);才把一个数和他的序号绑在一起了。
最后排一下序号，就可以找到每种数字的因子了，用一个动态数组存起来，所以需要先排序，不排序的话还得想办法标记。
ac代码

#include <bits/stdc++.h>
#define endl '\n'
using namespace std;
const int N = sqrt(1e7)+1;
vector<int>q,times,temp;
bool prime[N];
int len;
vector<int> p;
void prime2()
{for (int i = 2; i < N; i++) prime[i] = true;for (int i = 2; i < N; i++) {if (prime[i]) p.push_back(i);for (int j = 1; j*i <N ; j++) {prime[i*j]=false;}}len=p.size();
}
void asd(int x)//求出x的所有因子并存起来
{times.clear();int m,s=1;times.push_back(1);for (int i = 0; i < len && p[i] <= x; i++){m=1;if (x % p[i] == 0){while (x% p[i] == 0){x /= p[i];m*=p[i];q.push_back(m);}int length=times.size();int len2=q.size();for(int k=0;k<len2;k++){for(int j=0;j<length;j++)times.push_back(q[k]*times[j]);                 q.pop_back();}}}if(x>1)//大于sqrt（最开始的x）的质因子最多有一个，也就是剩下的x，这点也是刚想通，因为x已经无法被小于sqrt的质数分解了，所以剩下的x一定是质子；{int length=times.size();for(int j=0;j<length;j++)times.push_back(x*times[j]);                  }}
int main() {prime2();int n ; cin >> n;vector<pair<int, int> > arr(n), ans(n);for (int i = 0; i < n; i++) {scanf("%d",& arr[i].first);arr[i].second = i;}sort(arr.begin(), arr.end());for (int i = 0; i < n; i++) {if (i == 0 || arr[i-1].first != arr[i].first) {asd(arr[i].first);ans[arr[i].second] = make_pair(times[times.size()-1], arr[i].first/times[times.size()-1]);times.pop_back();} else {if(times.empty()){cout<<"NO"<<endl;return 0;}//说明p[i]的所有组合已经组完了，所以没法不重复的分解了ans[arr[i].second] =make_pair(times[times.size()-1], arr[i].first / times[times.size()-1]);times.pop_back(); }
}cout << "YES" << endl;for (int i = 0; i < n; i++) {printf("%d %d\n",ans[i].first ,ans[i].second );}return 0;
}在这里插入代码片

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