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Codeforces Round #838 (Div. 2)——B. Make Array Good

Problem - B - Codeforces

An array bb of mm positive integers is good if for all pairs ii and jj (1≤i,j≤m1≤i,j≤m), max(bi,bj)max(bi,bj) is divisible by min(bi,bj)min(bi,bj).

You are given an array aa of nn positive integers. You can perform the following operation:

Select an index ii (1≤i≤n1≤i≤n) and an integer xx (0≤x≤ai0≤x≤ai) and add xx to aiai, in other words, ai:=ai+xai:=ai+x.
After this operation, ai≤1018ai≤1018 should be satisfied.

You have to construct a sequence of at most nn operations that will make aa good. It can be proven that under the constraints of the problem, such a sequence of operations always exists.

Input

Each test contains multiple test cases. The first line contains a single integer tt (1≤t≤1041≤t≤104) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer nn (1≤n≤1051≤n≤105) — the length of the array aa.

The second line of each test case contains nn space-separated integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109) — representing the array aa.

It is guaranteed that the sum of nn over all test cases does not exceed 105105.

Output

For each test, output a single integer pp (0≤p≤n0≤p≤n) — denoting the number of operations in your solution.

In each of the following pp lines, output two space-separated integers — ii and xx.

You do not need to minimize the number of operations. It can be proven that a solution always exists.

Example

input

4
4
2 3 5 5
2
4 8
5
3 4 343 5 6
3
31 5 17

output

问题解析

题目是说给你n个数，你每次可以给一个数ai加上不大于aj的值，最多能进行n次操作，现在让你经过操作后让数组满足：选择任意两个位置i和j，满足i到j中的最大值可以整除最小值。

我们对数组排序，遍历数组，如果当前数直接是上一个数的倍数，我们不操作，如果当前数不是上一个数的倍数，那么很显然我们可以通过给当前数ai加上（ai-a(i-1))%a(i-1)来把ai变成上一个数的倍数。这样每个数最多只会操作一次。

AC代码

#include<iostream>
using namespace std;
#include<vector>
#include<algorithm>
#include<math.h>
#include<set>
#include <random>
#include<numeric>
#include<string>
#include<string.h>
#include<iterator>
#include<fstream>
#include<map>
#include<unordered_map>
#include<stack>
#include<list>
#include<queue>
#include<iomanip>
#include<bitset>

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)

#define endl '\n'
//#define int ll
#define PI acos(-1)
#define INF 0x3f3f3f3f
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll>PII;
const int N = 1e6 + 50, MOD = 1e9 + 7;

int a[N];
void solve()
{
    
    int n, mn = 1e9 + 10;;
    cin >> n;
    vector<PII>v(n);
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
        v[i - 1].first = a[i], v[i - 1].second = i;
        mn = min(mn, a[i]);
    }
    sort(v.begin(), v.end());
    cout << n << endl;
    for (int i = 0; i < n; i++)
    {
        if (v[i].first % mn == 0)
        {
            cout << v[i].second << " " << 0 << endl;
            mn = v[i].first;
        }
        else
        {
            cout << v[i].second << " " << mn - v[i].first % mn << endl;
            mn = v[i].first + (mn - v[i].first % mn);
        }
    }
}

signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int t = 1;

    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}