# A
# 题意
# 题解
1 //A 2 #include<bits/stdc++.h> 3 using namespace std; 4 void solve(int r){ 5 string str1,str2; 6 cin>>str1>>str2; 7 cout<<"Game #"<<r<<":"<<endl; 8 cout<<" "<<"The input words are "<<str1<<" and "<<str2<<"."<<endl; 9 reverse(str1.begin(),str1.end()); 10 reverse(str2.begin(),str2.end()); 11 string res1,res2; 12 for(int i=0,j=0;i<str1.size()&&j<str2.size();i++,j++){ 13 if(str1[i] != str2[j] || i==str1.size()-1 || j==str2.size()-1){ 14 res1=str1.substr(i,str1.size()); 15 res2=str2.substr(j,str2.size()); 16 break; 17 } 18 } 19 reverse(res1.begin(),res1.end()); 20 reverse(res2.begin(),res2.end()); 21 cout<<" The words entered in the notebook are "<<res1<<" and "<<res2<<"."<<endl; 22 cout<<endl; 23 } 24 int main(){ 25 int t; 26 cin>>t; 27 for(int i=1;i<=t;i++){ 28 solve(i); 29 } 30 }
# B
# 题意
# 题解
1 // B 2 #include<bits/stdc++.h> 3 #define pdd pair<double,double> 4 #define db double 5 #define x first 6 #define y second 7 using namespace std; 8 const int N=55; 9 pdd a[N]; 10 void solve(int r){ 11 cout<<"Test case #"<<r<<":"<<" "; 12 int res=0; 13 int n; 14 cin>>n; 15 for(int i=0;i<n;i++) 16 cin>>a[i].x>>a[i].y; 17 unordered_map<string,bool>rec; 18 for(int i=0;i<n;i++) 19 for(int j=0;j<i && j!=i;j++) 20 for(int k=0;k<j && k!=i && k!=j;k++){ 21 db x1 = a[i].x, y1 = a[i].y; 22 db x2 = a[j].x, y2 = a[j].y; 23 db x3 = a[k].x, y3 = a[k].y; 24 db d12 = sqrt(pow(x1 - x2, 2) + pow(y1 - y2, 2)); 25 db d13 = sqrt(pow(x1 - x3, 2) + pow(y1 - y3, 2)); 26 db d23 = sqrt(pow(x2 - x3, 2) + pow(y2 - y3, 2)); 27 db s=(d12+d23+d13)/2; 28 db delta = (s-d12)*(s-d13)*(s-d23); 29 db jud=fabs(x1*y2+y1*x3+x2*y3-y2*x3-y3*x1-y1*x2)/2; 30 if(delta <=0 || jud==0) 31 continue; 32 else { 33 res++; 34 } 35 } 36 37 cout<<res<<" "<<"triangle(s) can be formed."<<endl; 38 cout<<endl; 39 } 40 int main(){ 41 ios::sync_with_stdio(0); 42 cin.tie(0); 43 cout.tie(0); 44 int t; 45 cin>>t; 46 for(int i=1;i<=t;i++){ 47 solve(i); 48 } 49 }
# C
# 题意
# 题解
1 //C 2 #include<bits/stdc++.h> 3 #define pdd pair<double,double> 4 #define db double 5 #define fi first 6 #define se second 7 using namespace std; 8 const int N=55; 9 pdd a[N]; 10 unordered_map<int,string>u; 11 12 void solve(int r){ 13 string a; 14 unordered_map<char,int>p; 15 cin>>a; 16 cout<<"UCF Hold-em #"<<r<<": "<<a<<endl; 17 for(int i=0;i<a.size();i++){ 18 p[a[i]]++; 19 } 20 int res; 21 bool flag=0; 22 while(!flag){ 23 for(auto x:p){ 24 if(x.se == 4) { 25 res=5; 26 flag=1; 27 } 28 } 29 if(flag)break; 30 bool cnt1=0,cnt2=0; 31 for(auto x:p){ 32 if(x.se>=3) { 33 cnt1=1; 34 continue; 35 } 36 if(x.se>=2){ 37 cnt2=1; 38 continue; 39 } 40 } 41 if(cnt1 && cnt2){ 42 flag=1; 43 res=4; 44 } 45 if(flag)break; 46 for(auto x:p){ 47 if(x.se == 3) { 48 res=3; 49 flag=1; 50 } 51 } 52 if(flag)break; 53 for(auto x:p){ 54 if(x.se == 2) { 55 res=2; 56 flag=1; 57 } 58 } 59 if(flag)break; 60 res=1; 61 break; 62 } 63 cout<<"Best possible hand: "<<u[res]<<endl<<endl; 64 } 65 int main(){ 66 ios::sync_with_stdio(0); 67 cin.tie(0); 68 cout.tie(0); 69 int t; 70 cin>>t; 71 u[5]="FOUR OF A KIND"; 72 u[4]="FULL HOUSE"; 73 u[3]="THREE OF A KIND"; 74 u[2]="TWO OF A KIND"; 75 u[1]="BUST"; 76 for(int i=1;i<=t;i++){ 77 solve(i); 78 } 79 }
# D
# 题意
# 题解
# E
# 题意
# 题解
# F
# 题意
# 题解
# G
# 题意
# 题解
# H
# 题意
# 题解
# I
# 题意
# 题解
1 #include<queue> 2 #include<cassert> 3 #include<cstdio> 4 #include<cstring> 5 #include<cstdlib> 6 #include<iostream> 7 #include<algorithm> 8 #include<unordered_map> 9 #define fi first 10 #define se second 11 12 using namespace std; 13 14 typedef long long ll; 15 typedef pair<int,int>pii; 16 17 const int mod=1e9+7; 18 const int inf=0x3f3f3f3f; 19 const int N=1e6+5; 20 21 ll gcd(ll a,ll b){return b?gcd(b,a%b):a;} 22 ll qmi(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} 23 24 pii a[N]; 25 26 bool check(pii a,pii b,pii c){ 27 int x1=b.fi-a.fi,x2=c.fi-a.fi; 28 int y1=b.se-a.se,y2=c.se-a.se; 29 if(x1*y2==x2*y1) return false; 30 return true; 31 } 32 33 int main(){ 34 // freopen("in","r",stdin); 35 int CASE=0; 36 int _;for(scanf("%d",&_);_;_--){ 37 int n;scanf("%d",&n); 38 for(int i=1;i<=n;i++) scanf("%d%d",&a[i].fi,&a[i].se); 39 int ans=0; 40 for(int i=1;i<=n;i++) 41 for(int j=i+1;j<=n;j++) 42 for(int k=j+1;k<=n;k++) 43 ans+=check(a[i],a[j],a[k]); 44 printf("Test case #%d: %d triangle(s) can be formed.\n\n",++CASE,ans); 45 } 46 return 0; 47 }