Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Description
-
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 ai 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 ai: the number if teeth in the ith gear.
1 ≤ n ≤ 104
2 ≤ ai ≤ 106
Output
For each testcase print a single integer: the number of distinct pairs that satisfy the problem statement.
Sample Input
2 5 4 6 7 8 12 6 2 2 2 3 3 4
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 + ;
int a[MAXN];
typedef long long LL;
int main(){
int T, n;
scanf("%d", &T);
while(T--){
scanf("%d", &n);
for(int i = ; i < n; i++){
scanf("%d", a + i);
}
sort(a, a + n);
LL ans = ;
for(int i = ; i < n; i++){
if(a[i] == ){
ans += (n - i - );
}
else{
for(int j = ; j * a[i] <= a[n - ]; j++){
int r = upper_bound(a + i + , a + n, a[i]*j) - (a + i + );
int l = lower_bound(a + i + , a + n, a[i]*j) - (a + i + );
if(l == r)continue;
else{
ans += r - l;
}
}
}
}
printf("%lld\n", ans);
}
return ;
}