题意：
给你4*n个数，让你判断能不能用这个4*n个数为边凑成n个矩形，使的每个矩形面积相等
题解：
原本是想着用二分来找出来那个最终的面积，但是仔细想一想，那个面积只能是给出的4*n个数中的最小值和最大值的乘积，如果这两个长度不凑成一个矩形，那么肯定全部矩形的面积会出现不一致的
代码：
1 //The idea was to use dichotomies to find that area, and then use that area to figure it out, but it's more complicated than that
2 //If you don't use the smallest and largest as the two sides of a rectangle, there will always be an area larger than that
3 #include <stdio.h>
4 #include <algorithm>
5 #include <iostream>
6 #include <string.h>
7 using namespace std;
8 const int maxn = 405;
9 typedef long long ll;
10 int v[maxn],w[maxn];
11 int main()
12 {
13 int t;
14 scanf("%d",&t);
15 while(t--)
16 {
17 int n;
18 scanf("%d",&n);
19 for(int i=1;i<=4*n;++i)
20 {
21 scanf("%d",&v[i]);
22 }
23 sort(v+1,v+1+4*n);
24 int x=v[1]*v[4*n],flag=1;
25 for(int i=2;i<=2*n;++i)
26 {
27 if(v[i]*v[4*n-i+1]!=x)
28 {
29 flag=0;
30 break;
31 }
32 }
33 for(int i=2;i<=4*n;i+=2)
34 {
35 if(v[i]!=v[i-1])
36 {
37 flag=0;
38 break;
39 }
40 }
41 if(flag) printf("YES\n");
42 else printf("NO\n");
43 }
44 return 0;
45 }
题意：
给你n个数，让你找出来这n个数的公因数有多少个
题解：
肯定不能暴力for循环找，一个数的公因数的因数也是这个数的因数，那么我们就可以找出来这n个数的最大公共因数，然后再在这个因数里面找因数
代码：
1 #include <bits/stdc++.h>
2
3 #define ll long long
4
5 #define sc scanf
6
7 #define pr printf
8
9 using namespace std;
10
11 ll gcd(ll a, ll b)
12
13 {
14
15 return b == 0 ? a : gcd(b, a % b);
16
17 }
18
19 int main()
20
21 {
22
23 int n;
24
25 scanf("%d", &n);
26
27 ll t1, t2;
28
29 sc("%lld", &t1);
30
31 for (int i = 1; i < n; i++)
32
33 {
34
35 sc("%lld", &t2);
36
37 t1 = gcd(t1, t2);
38
39 }
40
41 ll qq = sqrt(t1);
42
43 ll ans = 0;
44
45 for (ll i = 1; i <= qq; i++)
46
47 {
48
49 if (t1 % i == 0)
50
51 {
52
53 ans++;
54
55 if (t1 / i != i)
56
57 ans++;
58
59 }
60
61 }
62
63 printf("%lld", ans);
64
65 }