2017-2018 ACM-ICPC Northern Eurasia (Northeastern European Regional) Contest (NEERC 17)（B,D）

https://codeforces.com/gym/101630/attachments

B - Box

根据样例可以得出几个式子，具体看代码，只要满足其中之一就可以

 1 #include
 2 using namespace std;
 3 typedef long long ll;
 4 const int N=1e3+5;
 5 ll a[100];
 6 int main(){
 7     ll w,h;
 8     for(int i=1;i<=3;i++)cin>>a[i];
 9     sort(a+1,a+4);
10     cin>>w>>h;
11     ll x=2*a[2]+2*a[1];
12     ll y=a[3]+2*a[1];
13     
14     if(x<y)swap(x,y);
15     ll p=a[2]+2*a[1];
16     ll q=2*a[1]+2*a[3];
17     if(p<q)swap(p,q);
18     if(w<h)swap(w,h); 
19     
20     ll u=a[2]+a[1];
21     ll v=3*a[3]+a[2]+a[1];
22     if(u<v)swap(u,v);
23     
24     ll g=a[3]+a[1];
25     ll k=3*a[2]+a[1]+a[3];
26     if(g<k)swap(g,k);
27     
28     
29     ll b=a[2]+a[3];
30     ll c=3*a[1]+a[2]+a[3];
31     if(b<c)swap(b,c);
32     int f=0;
33     if(w>=x&&h>=y)f++;
34     if(w>=p&&h>=q)f++;
35     if(w>=u&&h>=v)f++;
36     if(w>=g&&h>=k)f++;
37     if(w>=b&&h>=c)f++;
38     if(f)printf("Yes\n");
39     else printf("No\n");
40 }

D - Designing the Toy

这题当时不会写赛后看懂了某个博主的代码

主要是在一个面上构造，然后跟据输入顺序对这个面进行调整，构造方式结合下图和代码很容易看懂：

 

 

 1 #define bug(x) cout<<#x<<" is "< 2 #define IO std::ios::sync_with_stdio(0)
 3 #include 
 4 #define iter ::iterator
 5 #define pa pair
 6 #define pp pair
 7 using namespace  std;
 8 #define ll long long
 9 #define mk make_pair
10 #define pb push_back
11 #define se second
12 #define fi first
13 #define ls o<<1
14 #define rs o<<1|1
15 const ll mod=1e9+7;
16 const int N=1e3+10;
17 int a[N],b[N];
18 int mp[N][N];
19 struct node{
20     int x,y,z;
21     node(int x1,int y1,int z1):x(x1),y(y1),z(z1){}
22 };
23 vectorv;
24 void gao(int x,int y,int z){
25     for(int i=0;i){
26         v.pb(node(0,i,i));
27         mp[i][i]=1;
28     }
29     for(int i=x;i){
30         v.pb(node(0,x-1,i));
31         mp[x-1][i]=1;
32     }
33     int g=z-y;
34     for(int i=0;i){
35         for(int j=0;j){
36             if(!mp[i][j]&&g>0){
37                 v.pb(node(0,i,j));
38                 g--;
39             }
40         }
41     }
42     if(g){
43         cout<<-1<<endl;
44         return;
45     }
46     if(b[1]==a[1]&&b[2]==a[3]&&b[3]==a[2]){
47         for(int i=0;i){
48             swap(v[i].x,v[i].y);
49         }
50     }
51     else if(b[1]==a[3]&&b[2]==a[2]&&b[3]==a[1]){
52         for(int i=0;i){
53             swap(v[i].x,v[i].z);
54         }
55     }
56     else if(b[1]==a[2]&&b[2]==a[1]&&b[3]==a[3]){
57         for(int i=0;i){
58             swap(v[i].y,v[i].z);
59         }
60     }
61     else if(b[1]==a[2]&&b[2]==a[3]&&b[3]==a[1]){
62         for(int i=0;i){
63             swap(v[i].x,v[i].y);
64             swap(v[i].x,v[i].z);
65         }
66     }
67     else if(b[1]==a[3]&&b[2]==a[1]&&b[3]==a[2]){
68         for(int i=0;i){
69             swap(v[i].x,v[i].y);
70             swap(v[i].y,v[i].z);
71         }
72     }
73     printf("%d\n",v.size());
74     for(auto tmp:v){
75         printf("%d %d %d\n",tmp.x,tmp.y,tmp.z);
76     }
77 }   
78 int main(){
79     for(int i=1;i<=3;i++){
80         cin>>a[i];
81         b[i]=a[i];
82     }
83     sort(a+1,a+4);
84     gao(a[1],a[2],a[3]);
85 }