思路:首先对于1 2 3 4 . . . 2e5 我想啊,让 i=3 到 2e5-1的数全部 i/2e5 那么这些数就全为1了,1 2 1 1 1 … 1 1 2e5 ,我们这时操作次数还剩下8次(1,2,2e5没用掉),但2e5/2 8次不够用啊,所以接着我想留出一个10来现在操作次数剩下9次,2e5/10 用掉了6次,10/2又要用掉4次,比9次多了一次.那么我想这让2e5/a 用掉5次,a/2还是用4次就可以了。将a最大化16,16/2用4次,2e5/16用5次,那不就解决了嘛。但n<=16还需特殊对待。
Code:
#include<iostream>
#define FAST ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
const int Max = 1e6 + 5;
int main()
{
FAST;
int t;cin >> t;
while (t--)
{
int n;cin >> n;
cout << n + 5 << endl;
if (n <= 16)
{
for (int i = 3;i <= n - 1;i++) cout << i << " " << n << endl;
for (int i = 1;i <= 8;i++)cout << n << " " << 2 << endl;
}
else
{
for (int i = 3;i <= 15;i++)cout << i << " " << n << endl;
for (int i = 17;i <= n - 1;i++)cout << i << " " << n << endl;
for (int i = 1;i <= 5;i++)cout << n << " " << 16 << endl;
for (int i = 1;i <= 4;i++)cout << 16 << " " << 2 << endl;
}
}
}