What a Mess（二分）

最新推荐文章于 2025-08-14 22:07:01 发布

转载最新推荐文章于 2025-08-14 22:07:01 发布 · 105 阅读

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#数据结构与算法

本文介绍了一道算法题目——计数齿轮配对，通过枚举倍数与二分查找的方法来解决齿轮间的整除配对问题。文章提供了完整的代码实现，并解释了核心思路。

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What a Mess

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status Practice Gym 100947I

Description

standard input/output Announcement

Statements
Alex is a very clever boy, after all he is the son of the greatest watchmaker in Odin.

One day, Alex was playing with some old watches and he found n gears, each gear has a_i teeth in it. Alex likes to make beautiful pairs of gears, he thinks a pair of gears is beautiful if we attach the two gears together and spin the first gear exactly one rotation, then the other gear spins an integer number of rotations. For example a pair of 8 and 4 is beautiful, whereas a pair of 8 and 5 isn't, neither is pair of 4 and 8.

Now Alex is curious, he wants to know how many beautiful pairs are there. Counting is not Alex's thing, so he asked you to help him.

Input

The first line of input contains one integer T: The number of test cases you need to process.

Each test case consists of two lines. The first line is a single integer n: the number of gears Alex has. The second line contains n space separated integers a_i: the number if teeth in the i^th gear.

1 ≤ n ≤ 10⁴

2 ≤ a_i ≤ 10⁶

Output

For each testcase print a single integer: the number of distinct pairs that satisfy the problem statement.

Sample Input

Input

2 5 4 6 7 8 12 6 2 2 2 3 3 4

Output

3 7

Hint

note that we consider two pair distinct when they differ by at least one gear.

In the first sample the pairs are: (4,8) , (4,12) , (6,12

题解：问可以整除的对数是多少；枚举倍数，二分；

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int MAXN = 1e4 + 100;
int a[MAXN];
typedef long long LL;
int main(){
    int T, n;
    scanf("%d", &T);
    while(T--){
        scanf("%d", &n);
        for(int i = 0; i < n; i++){
            scanf("%d", a + i);
        }
        sort(a, a + n);
        LL ans = 0;
        for(int i = 0; i < n; i++){
            if(a[i] == 1){
                ans += (n - i - 1);
            }
            else{
                for(int j = 1; j * a[i] <= a[n - 1]; j++){
                    int r = upper_bound(a + i + 1, a + n, a[i]*j) - (a + i + 1);
                    int l = lower_bound(a + i + 1, a + n, a[i]*j) - (a + i + 1);
                    if(l == r)continue;
                    else{
                        ans += r - l;
                    }
                }
            }
        }
        printf("%lld\n", ans);
    }
    return 0;
}