Codeforces Round #728 (Div. 2)

文章目录

A. Pretty Permutations
B. Pleasant Pairs
C. Great Graphs

A. Pretty Permutations

There are nnn cats in a line, labeled from 111 to nnn, with the iii-th cat at position iii. They are bored of gyrating in the same spot all day, so they want to reorder themselves such that no cat is in the same place as before. They are also lazy, so they want to minimize the total distance they move. Help them decide what cat should be at each location after the reordering.

For example, if there are 333 cats, this is a valid reordering: [3,1,2][3, 1, 2][3,1,2]. No cat is in its original position. The total distance the cats move is 1+1+2=41 + 1 + 2 = 41+1+2=4 as cat 111 moves one place to the right, cat 222 moves one place to the right, and cat 333 moves two places to the left.

Input

The first line contains a single integer ttt (1≤t≤1001 \leq t \leq 1001≤t≤100) — the number of test cases. Then ttt test cases follow.

The first and only line of each test case contains one integer nnn (2≤n≤1002 \leq n \leq 1002≤n≤100) — the number of cats.

It can be proven that under the constraints of the problem, an answer always exist.

Output

Output ttt answers, one for each test case. Each answer consists of nnn integers — a permutation with the minimum total distance. If there are multiple answers, print any.

Example

input

2
2
3

output

2 1
3 1 2

Note

For the first test case, there is only one possible permutation that satisfies the conditions: [2,1][2, 1][2,1].

The second test case was described in the statement. Another possible answer is [2,3,1][2, 3, 1][2,3,1].

#include <iostream>
#include <algorithm>
using namespace std;int main(){int t;cin >> t;while(t--){int n;cin >> n;if(n%2==0){for(int i=1; i<n; ){cout << i+1 <<" " << i <<" ";i += 2;}cout << endl;}else{for(int i=1; i<=n-3; ){cout << i+1 <<" " << i <<" ";i += 2;}cout << n << " " << n-2 << " " << n-1 << " " ;cout << endl;}}return 0;
}

B. Pleasant Pairs

You are given an array a1,a2,…,ana_1, a_2, \dots, a_na1,a2,…,an consisting of nnn distinct integers. Count the number of pairs of indices (i,j)(i, j)(i,j) such that i<ji < ji<j and ai⋅aj=i+ja_i \cdot a_j = i + jai⋅aj=i+j.

Input

The first line contains one integer ttt (1≤t≤1041 \leq t \leq 10^41≤t≤104) — the number of test cases. Then ttt cases follow.

The first line of each test case contains one integer nnn (2≤n≤1052 \leq n \leq 10^52≤n≤105) — the length of array aaa.

The second line of each test case contains nnn space separated integers a1,a2,…,ana_1, a_2, \ldots, a_na1,a2,…,an (1≤ai≤2⋅n1 \leq a_i \leq 2 \cdot n1≤ai≤2⋅n) — the array aaa. It is guaranteed that all elements are distinct.

It is guaranteed that the sum of nnn over all test cases does not exceed 2⋅1052 \cdot 10^52⋅105.

Output

For each test case, output the number of pairs of indices (i,j)(i, j)(i,j) such that i<ji < ji<j and ai⋅aj=i+ja_i \cdot a_j = i + jai⋅aj=i+j.

Example

input

output

1
1
3

Note

For the first test case, the only pair that satisfies the constraints is (1,2)(1, 2)(1,2), as a1⋅a2=1+2=3a_1 \cdot a_2 = 1 + 2 = 3a1⋅a2=1+2=3

For the second test case, the only pair that satisfies the constraints is (2,3)(2, 3)(2,3).

For the third test case, the pairs that satisfy the constraints are (1,2)(1, 2)(1,2), (1,5)(1, 5)(1,5), and (2,3)(2, 3)(2,3).

分析

a[i] * a[j] ，根据 a[i] 来找 a[j] ， i + j 一定是 a[i] 的倍数，i 是不变的，所以 j 的增加量一定是a[i] 的倍数。

#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 2e5 + 5;
ll a[N];int main(){int T;cin >> T;while(T--){int n;cin >> n;ll ans = 0;for(int i = 1; i <= n; i++)cin >> a[i];for(int i = 1; i <= n; i++)for(int j = a[i] - i; j <= n; j += a[i]){if(j <= i)continue;if(a[i] * a[j] == i + j)ans++;}cout << ans << endl;}return 0;}

C. Great Graphs

Farmer John has a farm that consists of nnn pastures connected by one-directional roads. Each road has a weight, representing the time it takes to go from the start to the end of the road. The roads could have negative weight, where the cows go so fast that they go back in time! However, Farmer John guarantees that it is impossible for the cows to get stuck in a time loop, where they can infinitely go back in time by traveling across a sequence of roads. Also, each pair of pastures is connected by at most one road in each direction.

Unfortunately, Farmer John lost the map of the farm. All he remembers is an array ddd, where did_idi is the smallest amount of time it took the cows to reach the iii-th pasture from pasture 111 using a sequence of roads. The cost of his farm is the sum of the weights of each of the roads, and Farmer John needs to know the minimal cost of a farm that is consistent with his memory.

Input

The first line contains one integer ttt (1≤t≤1041 \leq t \leq 10^41≤t≤104) — the number of test cases. Then ttt cases follow.

The first line of each test case contains a single integer nnn (1≤n≤1051 \leq n \leq 10^51≤n≤105) — the number of pastures.

The second line of each test case contains nnn space separated integers d1,d2,…,dnd_1, d_2, \ldots, d_nd1,d2,…,dn (0≤di≤1090 \leq d_i \leq 10^90≤di≤109) — the array ddd. It is guaranteed that d1=0d_1 = 0d1=0.

It is guaranteed that the sum of nnn over all test cases does not exceed 10510^5105.

Output

For each test case, output the minimum possible cost of a farm that is consistent with Farmer John’s memory.

Example

input

3
3
0 2 3
2
0 1000000000
1
0

output

-3
0
0

Note

In the first test case, you can add roads

from pasture 111 to pasture 222 with a time of 222,
from pasture 222 to pasture 333 with a time of 111,
from pasture 333 to pasture 111 with a time of −3-3−3,
from pasture 333 to pasture 222 with a time of −1-1−1,
from pasture 222 to pasture 111 with a time of −2-2−2.

The total cost is 2+1+−3+−1+−2=−32 + 1 + -3 + -1 + -2 = -32+1+−3+−1+−2=−3.

In the second test case, you can add a road from pasture 111 to pasture 222 with cost 100000000010000000001000000000 and a road from pasture 222 to pasture 111 with cost −1000000000-1000000000−1000000000. The total cost is 1000000000+−1000000000=01000000000 + -1000000000 = 01000000000+−1000000000=0.

In the third test case, you can’t add any roads. The total cost is 000.

分析

相当于负向的边全部加上，正向只在相邻pasture加边

#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;const int N = 1E5 + 10;
int a[N];
ll sum,ans;int main(){int T;cin >> T;while(T--){int n;cin >> n;sum = ans = 0;for(int i=0; i<n; i++){cin >> a[i];sum += a[i];}sort(a, a+n); // 由小到大排序 ans = sum;for(int i=1; i<n-1; i++){sum -= 1ll*(n-i)*(a[i]-a[i-1]);ans += sum;}ans -= a[n-1];cout << -ans <<endl; }return 0;
}