[贪心]Codeforces Equal Rectangles
You are given 4n4n sticks, the length of the ii-th stick is aiai.
You have to create nn rectangles, each rectangle will consist of exactly 44 sticks from the given set. The rectangle consists of four sides, opposite sides should have equal length and all angles in it should be right. Note that each stick can be used in only one rectangle. Each stick should be used as a side, you cannot break the stick or use it not to the full length.
You want to all rectangles to have equal area. The area of the rectangle with sides aa and bb is a⋅ba⋅b.
Your task is to say if it is possible to create exactly nn rectangles of equal area or not.
You have to answer qq independent queries.
The first line of the input contains one integer qq (1≤q≤5001≤q≤500) — the number of queries. Then qq queries follow.
The first line of the query contains one integer nn (1≤n≤1001≤n≤100) — the number of rectangles.
The second line of the query contains 4n4n integers a1,a2,…,a4na1,a2,…,a4n (1≤ai≤1041≤ai≤104), where aiaiis the length of the ii-th stick.
For each query print the answer to it. If it is impossible to create exactly nn rectangles of equal area using given sticks, print "NO". Otherwise print "YES".
5 1 1 1 10 10 2 10 5 2 10 1 1 2 5 2 10 5 1 10 5 1 1 1 2 1 1 1 1 1 1 1 1 1 10000 10000 10000 10000
YES YES NO YES YES
题意:
给你4*n个棍子，问能否用这些棍子组成n个面积相同的矩形
思路:
给n个棍子按大小排序，并统计其数量，第一次用最大和最小的棍子组成矩形，每次也是取出最大和最小的棍子组成矩形来判断是否和原矩形同面积，注意如果棍子用完了就要删掉这种棍子，如果棍子不够或多了就是不合法的要跳出循环
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int amn=1e4+5; 4 int cnt[amn]; 5 set<int> s; 6 set<int>::iterator it; 7 set<int>::iterator rit; 8 int main(){ 9 int q,n,in; 10 cin>>q; 11 while(q--){ 12 memset(cnt,0,sizeof cnt); 13 cin>>n; 14 s.clear(); 15 for(int i=1;i<=4*n;i++){ 16 cin>>in; 17 s.insert(in); 18 cnt[in]++; 19 } 20 it=s.begin(),rit=s.end();rit--; 21 int a=*it,b=*rit; 22 int need=(a)*(b),num=0; 23 bool valid=1; 24 if(cnt[a]&&cnt[b]&&cnt[a]%2==0&&cnt[b]%2==0){ 25 cnt[a]-=2;cnt[b]-=2; 26 num++; 27 if(cnt[a]==0) 28 s.erase(it++); 29 else if(cnt[a]<0){ 30 valid=0; 31 break; 32 } 33 if(a!=b&&cnt[b]==0) 34 s.erase(rit--); 35 else if(cnt[b]<0){ 36 valid=0; 37 break; 38 } 39 while(it!=s.end()&&s.size()){ 40 a=*it,b=*rit; 41 int now=(a)*(b); 42 if(now==need&&num<n){ 43 if(cnt[a]&&cnt[b]&&cnt[a]%2==0&&cnt[b]%2==0){ 44 cnt[a]-=2;cnt[b]-=2; 45 num++; 46 if(cnt[a]==0) 47 s.erase(it++); 48 else if(cnt[a]<0){ 49 valid=0; 50 break; 51 } 52 if(a!=b&&cnt[b]==0) 53 s.erase(rit--); 54 else if(cnt[b]<0){ 55 valid=0; 56 break; 57 } 58 } 59 else{ 60 valid=0; 61 break; 62 } 63 } 64 else{ 65 valid=0; 66 break; 67 } 68 } 69 } 70 else{ 71 valid=0; 72 } 73 if(valid&&num==n)cout<<"YES\n"; 74 else cout<<"NO\n"; 75 } 76 } 77 /*** 78 给你4*n个棍子，问能否用这些棍子组成n个面积相同的矩形 79 给n个棍子按大小排序，并统计其数量，第一次用最大和最小的棍子组成矩形，每次也是取出最大和最小的棍子组成矩形来判断是否和原矩形同面积，注意如果棍子用完了就要删掉这种棍子，如果棍子不够或多了就是不合法的要跳出循环 80 ***/
