Codeforces Round #579 (Div. 3) B Equal Rectangles、C. Common Divisors

B Equal Rectangles

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 }

C. Common Divisors

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 }