week3
SMU Winter 2023 Round #5 (Div.2)
Lucky?
int main(){ int t,a[6]; cin>>t; while(t--){ scanf("%1d%1d%1d%1d%1d%1d",&a[0],&a[1],&a[2],&a[3],&a[4],&a[5]); if(a[0]+a[1]+a[2]==a[3]+a[4]+a[5])cout<<"YES\n"; else cout<<"NO\n"; } return 0; }
Equal Candies
思路：找出最小的数
int main(){ int t,n,a[55]; cin>>t; while(t--){ int mi=N; cin>>n; for(int i=0;i<n;++i){ cin>>a[i]; if(a[i]<mi)mi=a[i]; } long long s=0; for(int i=0;i<n;++i){ s+=(a[i]-mi); } cout<<s<<'\n'; } return 0; }
Most Similar Words
思路：枚举每一对
#include<bits/stdc++.h> using namespace std; const int N=1e7+10; int main(){ int t,n,m; string a[55]; cin>>t; while(t--){ cin>>n>>m; for(int i=0;i<n;++i)cin>>a[i]; int mi=N; for(int i=0;i<n-1;++i) for(int j=i+1;j<n;++j){ int s=0; for(int k=0;k<m;++k){ s+=(a[i][k]+a[j][k]-2*min(a[i][k],a[j][k])); } mi=min(mi,s); } cout<<mi<<'\n'; } return 0; }
X-Sum
思路：枚举每个点的对角线和
#include<bits/stdc++.h> using namespace std; const int N=1e7+10; int main(){ int t,n,m; int a[205][205]; int dx[4]={-1,-1,1,1},dy[4]={-1,1,-1,1}; cin>>t; while(t--){ cin>>n>>m; int res=0; for(int i=1;i<=n;++i) for(int j=1;j<=m;++j) cin>>a[i][j]; for(int i=1;i<=n;++i) for(int j=1;j<=m;++j){ int s=0; s+=a[i][j]; for(int k=1;k<=max(n,m);++k){ for(int u=0;u<4;++u){ int xx=k*dx[u]+i,yy=k*dy[u]+j; if(xx>0&&xx<=n&&yy>0&&yy<=m) s+=a[xx][yy]; } } res=max(res,s); } cout<<res<<'\n'; } return 0; }
Eating Queries
思路：对所有数降序排序后求前缀和，前缀和为升序，二分查找x
#include<bits/stdc++.h> using namespace std; const int N=150005; typedef long long ll; bool cmp(int a,int b){ return a>b; } int main(){ int t,n,q,a[N]; ll x; cin>>t; while(t--){ cin>>n>>q; for(int i=0;i<n;++i)cin>>a[i]; sort(a,a+n,cmp); for(int i=1;i<n;++i)a[i]+=a[i-1]; while(q--){ cin>>x; if(x>a[n-1])cout<<-1<<'\n'; else if(x==a[n-1])cout<<n<<'\n'; else{ ll s=lower_bound(a,a+n,x)-a; if(s<n)cout<<s+1<<'\n'; else cout<<-1<<'\n'; } } } return 0; }
Longest Strike
思路：用map存储每个数个数，刚好得到排列后的序列，查找个数都大于等于k的连续区间
#include<bits/stdc++.h> using namespace std; const int N=1e6+10; typedef long long ll; int t,n,k,l,r,maxl,mi,ma,len,last,a[N]; map<int,int>mp; int main(){ cin>>t; while(t--){ cin>>n>>k; mp.clear(),maxl=0,mi=0x3f3f3f3f,ma=0,len=0; for(int i=1;i<=n;++i){ cin>>a[i]; mp[a[i]]++; mi=min(mi,a[i]),ma=max(ma,a[i]); } last=mi-1; for(auto v:mp){ int i=v.first; if(i==ma&&v.second>=k&&i==last+1&&len+1>maxl)maxl=len+1,l=i-len,r=i; if(v.second>=k&&i==last+1)len++; else{ if(len>maxl)maxl=len,l=last+1-len,r=last; if(v.second>=k)len=1; else len=0; } last=i; } if(maxl!=0)cout<<l<<' '<<r<<'\n'; else if(mp[ma]>=k)cout<<ma<<' '<<ma<<'\n'; else cout<<"-1\n"; } return 0; }
White-Black Balanced Subtrees
思路：dfs求子树的点的黑白个数是否相同
#include<bits/stdc++.h> using namespace std; const int N=4010; vector<int>ve[N]; int f[N][2]; string s; void dfs(int u,int fa) { if(s[u]=='W') f[u][0]++; else f[u][1]++; for(int i=0;i<ve[u].size();i++) { int j=ve[u][i]; if(j==fa) continue; dfs(j,u); f[u][0]+=f[j][0],f[u][1]+=f[j][1]; } } int main(void) { int t; cin>>t; while(t--) { int n; cin>>n; for(int i=1;i<=n;i++) ve[i].clear(),f[i][0]=0,f[i][1]=0; for(int i=2;i<=n;i++) { int x; cin>>x; ve[x].push_back(i),ve[i].push_back(x); } cin>>s; s="0"+s; dfs(1,-1); int cnt=0; for(int i=1;i<=n;i++) if(f[i][0]==f[i][1]) cnt++; cout<<cnt<<'\n'; } return 0; }
思路：求i<j，a[i]>=a[j]的个数，easy：双指针，hard：树状数组或归并排序
#include<bits/stdc++.h> using namespace std; const int N=1e5*2+10; typedef long long ll; int t,n,a[N],tr[N]; int lowbit(int x){ return x&-x; } void update(int x,ll v){ for(int i=x;i<N;i+=lowbit(i))tr[i]+=v; } ll q(int x){ ll sum=0; for(int i=x;i;i-=lowbit(i))sum+=tr[i]; return sum; } int main(){ cin>>t; while(t--){ cin>>n; for(int i=0;i<=n+10;++i)tr[i]=0; for(int i=1;i<=n;++i)cin>>a[i]; ll ans=0; for(int i=1;i<=n;++i){ ans+=q(n+1)-q(a[i]-1); update(a[i],1); } cout<<ans<<'\n'; } return 0; }
SMU Winter 2023 Round #6 (Div.2)
OP
#include<bits/stdc++.h> using namespace std; int main(){ cout<<"fengqibisheng, yingyueerlai!"; return 0; }
Steel of Heart
思路：模拟每一种技能
#include<bits/stdc++.h> using namespace std; int main(){ int h1,h2; int m; cin>>h1>>h2>>m; char x; bool ok=false; int mm,ss,a,b,t[2][5]={{0,0,0,0,0},{0,0,0,0,0}}; while(m--){ cin>>mm>>x>>ss>>a; if(a==1){ h1+=800; ok=true; } else if(a==2){ h1+=h2; } else{ cin>>b; if(ok){ if(mm==t[0][b-1]&&ss>=t[1][b-1]||mm>t[0][b-1]){ h1+=double((125.0+0.06*h1))*0.1; t[1][b-1]=(ss+30)%60; if((ss+30)>=60)t[0][b-1]=mm+1; else t[0][b-1]=mm; } } } } cout<<h1; return 0; }
思路：排序后，若a[j]-a[i]=9 (i<j)，则少一种可能
#include<bits/stdc++.h> using namespace std; int main(){ const int N=500005; int n; cin>>n; vector<int>a(n); int s=n; for(int i=0;i<n;++i){ cin>>a[i]; } sort(a.begin(),a.end()); for(int i=0;i<n;++i) for(int j=i+1;j<n;++j){ if(a[j]-a[i]==9)s--; if(a[j]-a[i]>9)break; } cout<<s; return 0; }
Survivor
思路：计算出每个人存活需要的次数，排序后从小到大取
#include<bits/stdc++.h> using namespace std; const int N=2e5+5; int a[N],b[N],c[N],s[N]; int main(){ int n,k,m; cin>>n>>m>>k; for(int i=0;i<n;++i){ scanf("%d",a+i); } for(int i=0;i<n;++i){ scanf("%d",b+i); } for(int i=0;i<n;++i){ scanf("%d",c+i); } for(int i=0;i<n;++i){ long long all=a[i]-m*b[i]; if(all<=0){ int t=1; while(all+t*c[i]<=0)t++; s[i]=t; } } int cnt=0; sort(s,s+n); for(int i=0;i<n;++i){ cnt+=s[i]; if(cnt>k){ cout<<i; break; } } return 0; }
Simple Game
思路：判断总和的奇偶性
#include <bits/stdc++.h> using namespace std; int main(){ long long n,k; long long s=0; cin>>n; for(int i=1;i<=n;i++){ scanf("%lld",&k); s+=k; } if(s%2)printf("Alice"); else printf("Bob"); return 0; }
Bit
思路：记录n次操作分别对二进制第i位为0或1后的结果，对于每次询问，二进制从最高位到低位枚举，第i位当初始为0，结果为1或0的情况答案都为0；当初始为1，结果为1，且该情况的数不超过r，答案为1，其他都为0
#include<bits/stdc++.h> using namespace std; const int N=2e5+5; int n,q,a,b,x=0,y,xx[30],yy[30]; int main(){ cin>>n>>q; for(int i=0;i<30;++i)y|=(1<<i); for(int i=0;i<n;++i){ cin>>a>>b; if(a==1)x=x&b,y=y&b; else if(a==2)x=x|b,y=y|b; else x=x^b,y=y^b; } for(int i=29;i>=0;--i){ xx[i]=(x>>i)&1; yy[i]=(y>>i)&1; } int s; while(q--){ s=0; cin>>a; for(int i=29;i>=0;--i){ if(xx[i]!=1&&yy[i]==1){ if((s+(1<<i))<=a)s|=(1<<i); } } cout<<s<<'\n'; } return 0; }
Elden Ring
思路：对2n个数编号，若编号是m的倍数，该编号与编号%n交换
#include<bits/stdc++.h> using namespace std; int main(){ int n,m,k; cin>>n>>m>>k; int a[2005]; for(int i=1;i<=2*n;++i)a[i]=i; for(int i=1;i<=k;++i){ int j=i%n; if(j==0)j=n; if(i%m==0)swap(a[j],a[j+n]); } for(int i=1;i<=2*n;++i)cout<<a[i]<<' '; return 0; }
SMU Winter 2023 Round #7 (Div.2)
思路：统计每个字符数量，统计偶数对个数，若出现奇数，总数加一，若没有奇数，总数减一
#include<bits/stdc++.h> using namespace std; const int N=2e5+5; int main(){ int t,n; cin>>t; char a; while(t--){ int b[60]={0},s=0; cin>>n; for(int i=0;i<n;++i){ cin>>a; b[a-'A']++; } bool ok=false; for(int i=0;i<58;++i){ if(b[i]%2==1)ok=true; if(b[i]!=0) s+=((b[i]/2)*2); } if(ok)s++; cout<<s<<'\n'; } return 0; }
7 的意志
思路：用前缀和统计区间和为7777的个数
#include<bits/stdc++.h> using namespace std; const int N=1e5+5; int n,t,a[N]; int main(){ cin>>t; while(t--){ scanf("%d",&n); a[0]=0; for(int i=1;i<=n;++i)scanf("%d",&a[i]); int cnt=0; for(int i=1;i<=n;++i)a[i]+=a[i-1]; for(int i=1;i<=n;++i){ int s=a[i-1]+7777; int p= lower_bound(a+1,a+n+1,s)-a; if(a[p]==s)cnt++; } printf("%d\n",cnt); } return 0; }
思路：更新每一次操作后的物体的坐标，自己的位置，拿到的物体数
#include<bits/stdc++.h> using namespace std; const int N=1e5+5; int n,x[2005][2005],y[2005][2005],q,s[2005]; int px[2005],py[2005],ans[2005],a[2005],b[2005],op[2005],vis[2005][2005]; int main(){ cin>>n; for(int i=1;i<=n;++i){ cin>>x[0][i]>>y[0][i]; vis[0][i]=0; } cin>>q; for(int i=1;i<=q;++i){ cin>>op[i]; if(op[i]==1){ string ss; cin>>ss; if(ss[0]=='X')a[i]=1; else a[i]=2; cin>>b[i]; } else if(op[i]==2){ string ss; cin>>ss; if(ss[0]=='X')a[i]=1; else a[i]=2; } else if(op[i]==4)cin>>a[i]; } px[0]=py[0]=ans[0]=0; for(int i=1;i<=q;++i){ ans[i]=ans[i-1]; px[i]=px[i-1]; py[i]=py[i-1]; if(op[i]==4){ memcpy(x[i],x[a[i]-1],sizeof(x[i])); memcpy(y[i],y[a[i]-1],sizeof(y[i])); memcpy(vis[i],vis[a[i]-1],sizeof(vis[i])); ans[i]=ans[a[i]-1]; px[i]=px[a[i]-1]; py[i]=py[a[i]-1]; continue; } memcpy(x[i],x[i-1],sizeof (x[i])); memcpy(y[i],y[i-1],sizeof(y[i])); memcpy(vis[i],vis[i-1],sizeof(vis[i])); if(op[i]==1){ if(a[i]==1){ int l=min(px[i],px[i]+b[i]); int r=max(px[i],px[i]+b[i]); px[i]+=b[i]; if(px[i]!=0&&py[i]!=0){ px[i]=py[i]=0; continue; } for(int j=1;j<=n;++j){ if(!vis[i][j]&&y[i][j]==0&&l<=x[i][j]&&x[i][j]<=r){ vis[i][j]=1;++ans[i]; } } } else{ int l=min(py[i],py[i]+b[i]); int r=max(py[i],py[i]+b[i]); py[i]+=b[i]; if(px[i]!=0&&py[i]!=0){ px[i]=py[i]=0; continue; } for(int j=1;j<=n;++j){ if(!vis[i][j]&&x[i][j]==0&&l<=y[i][j]&&y[i][j]<=r){ vis[i][j]=1;++ans[i]; } } } } else if(op[i]==2){ if(a[i]==1){ for(int j=1;j<=n;++j)x[i][j]*=(-1); } else for(int j=0;j<=n;++j)y[i][j]*=(-1); } else if(op[i]==3){ for(int j=1;j<=n;++j)swap(x[i][j],y[i][j]); } for(int j=1;j<=n;++j) if(!vis[i][j]&&px[i]==x[i][j]&&py[i]==y[i][j]){ ++ans[i];vis[i][j]=1; break; } } cout<<ans[q]; return 0; }
转动命运之轮
思路：答案为快乐值总和与操作数的乘积，模拟n = 1 , 2 , ⋯ , 5 ，发现系数p 的值分别等于1 , 3 , 12 , 60 , 360
#include<bits/stdc++.h> using namespace std; const int N=998244353; typedef long long ll; ll n,sum=0,tmp=1; ll a[2005]; int main(){ cin>>n; for(int i=1;i<=n;++i){ cin>>a[i]; sum=(sum+a[i])%N; } for(int i=2;i<=n;++i) tmp=(tmp*(i+1))%N; cout<<(sum*tmp)%N; return 0; }
计算最小值
思路：对选取数的操作后使得和最小，就是让选取数中最大值减去最小值的值最小。选取数量为n，将所有数排列，从最小的开始，用双指针算法，当选取的数刚好为n时记录最小的res
#include<bits/stdc++.h> using namespace std; typedef pair<int,int>PII; const int N=998244353; typedef long long ll; int n,m; int read(){ int x = 0 , ch = getchar(); while( ch < '0' || ch > '9' ) ch = getchar(); while( ch >= '0' && ch <= '9' ) x = ( x << 3 ) + ( x << 1 ) + ch - '0' , ch = getchar(); return x; } vector<PII>v; int main(){ n=read(),m=read(); vector<int>cnt(n+1); for (int i = 1; i <= n; ++i) { for (int j = 1,x; j <= m; ++j) { x=read();v.push_back({x,i}); } } sort(v.begin(),v.end()); int res=1e9; for(int l=0,r=-1;l<v.size();++l){ while(r+1<v.size()&&cnt[0]!=n){ r++,cnt[v[r].second]++; if(cnt[v[r].second]==1)cnt[0]++; } if(cnt[0]!=n)break; if(v[l].second!=v[r].second)res=min(res,v[r].first-v[l].first); cnt[v[l].second]--; if(cnt[v[l].second]==0)cnt[0]--; } cout<<res<<'\n'; return 0; }
月光奏鸣曲
思路：将正方形一与旋转1次和2次的正方形二作比较
#include<bits/stdc++.h> using namespace std; const int N=1e5+5; int t,n,a[25][25],b[25][25]; int main(){ cin>>t; while(t--){ cin>>n; for(int i=0;i<n;++i) for(int j=0;j<n;++j)cin>>a[i][j]; for(int i=0;i<n;++i) for(int j=0;j<n;++j)cin>>b[i][j]; bool ok=true; int ans=0; for(int i=0;i<n&&ok;++i) for(int j=0;j<n;++j) if(a[i][j]!=b[i][j]){ ok=false; break; } if(!ok){ ok=true;ans=1; for(int i=0;i<n&&ok;++i) for(int j=0;j<n;++j) if(a[i][j]!=b[j][n-i-1]){ ok=false; break; } } if(!ok){ ok=true;ans=2; for(int i=0;i<n&&ok;++i) for(int j=0;j<n;++j)if(a[i][j]!=b[n-i-1][n-j-1]){ ok=false; break; } } if(!ok){ ok=true;ans=1; for(int i=0;i<n&&ok;++i) for(int j=0;j<n&&ok;++j)if(a[i][j]!=b[n-j-1][i]){ ok=false; break; } } if(ok)cout<<ans<<'\n'; else cout<<-1<<'\n'; } return 0; }
电子表校对
思路：将一个数字分段为三段，区分0~9
#include<bits/stdc++.h> using namespace std; const int N=1e5+5; int main(){ string a[5],b[5]; for(int i=0;i<3;++i){ getline(cin,a[i]); } for(int i=0;i<3;++i)getline(cin,b[i]); int c[6],d[6]; for(int i=0;i<6;++i){ string x[5]; for(int j=0;j<3;++j){ x[j]=a[j].substr(i*3,3); } if(x[0]==" "&&x[1]==" |"&&x[2]==" |")c[i]=1; else if(x[0]==" _ "&&x[1]==" _|"&&x[2]=="|_ ")c[i]=2; else if(x[0]==" _ "&&x[1]==" _|"&&x[2]==" _|")c[i]=3; else if(x[0]==" "&&x[1]=="|_|"&&x[2]==" |")c[i]=4; else if(x[0]==" _ "&&x[1]=="|_ "&&x[2]==" _|")c[i]=5; else if(x[0]==" _ "&&x[1]=="|_ "&&x[2]=="|_|")c[i]=6; else if(x[0]==" _ "&&x[1]==" |"&&x[2]==" |")c[i]=7; else if(x[0]==" _ "&&x[1]=="|_|"&&x[2]=="|_|")c[i]=8; else if(x[0]==" _ "&&x[1]=="|_|"&&x[2]==" _|")c[i]=9; else c[i]=0; for(int j=0;j<3;++j)x[j]=b[j].substr(i*3,3); if(x[0]==" "&&x[1]==" |"&&x[2]==" |")d[i]=1; else if(x[0]==" _ "&&x[1]==" _|"&&x[2]=="|_ ")d[i]=2; else if(x[0]==" _ "&&x[1]==" _|"&&x[2]==" _|")d[i]=3; else if(x[0]==" "&&x[1]=="|_|"&&x[2]==" |")d[i]=4; else if(x[0]==" _ "&&x[1]=="|_ "&&x[2]==" _|")d[i]=5; else if(x[0]==" _ "&&x[1]=="|_ "&&x[2]=="|_|")d[i]=6; else if(x[0]==" _ "&&x[1]==" |"&&x[2]==" |")d[i]=7; else if(x[0]==" _ "&&x[1]=="|_|"&&x[2]=="|_|")d[i]=8; else if(x[0]==" _ "&&x[1]=="|_|"&&x[2]==" _|")d[i]=9; else d[i]=0; } int h[2],m[2],s[2],hh,mm,ss; h[0]=c[0]*10+c[1],h[1]=d[0]*10+d[1]; m[0]=c[2]*10+c[3],m[1]=d[2]*10+d[3]; s[0]=c[4]*10+c[5],s[1]=d[4]*10+d[5]; if(h[0]==h[1]&&m[0]==m[1]&&s[0]==s[1]){ cout<<"gang gang hao\n"; hh=mm=ss=0; } else if((h[0]==h[1]&&((m[0]==m[1]&&s[0]>s[1])||(m[0]>m[1])))||h[0]>h[1]){ cout<<"early\n"; if(s[0]>=s[1])ss=s[0]-s[1]; else{ ss=60+s[0]-s[1];m[0]--; } if(m[0]>=m[1])mm=m[0]-m[1]; else{ mm=60+m[0]-m[1];h[0]--; } hh=h[0]-h[1]; } else { cout<<"late\n"; if(s[1]>=s[0])ss=s[1]-s[0]; else{ss=60+s[1]-s[0];m[1]--;} if(m[1]>=m[0])mm=m[1]-m[0]; else{mm=60+m[1]-m[0];h[1]--;} hh=h[1]-h[0]; } c[0]=hh/10,c[1]=hh%10,c[2]=mm/10,c[3]=mm%10,c[4]=ss/10,c[5]=ss%10; for(int i=0;i<3;++i){ for(int j=0;j<6;++j){ if(i==0){ if(c[j]==4||c[j]==1)cout<<" "; else cout<<" _ "; } else if(i==1){ if(c[j]==1||c[j]==7)cout<<" |"; else if(c[j]==2||c[j]==3)cout<<" _|"; else if(c[j]==5||c[j]==6)cout<<"|_ "; else if(c[j]==8||c[j]==9||c[j]==4)cout<<"|_|"; else cout<<"| |"; } else{ if(c[j]==1||c[j]==4||c[j]==7)cout<<" |"; else if(c[j]==2)cout<<"|_ "; else if(c[j]==3||c[j]==5||c[j]==9)cout<<" _|"; else cout<<"|_|"; } }cout<<'\n'; } return 0; }
第二版本：
#include<bits/stdc++.h> using namespace std; typedef pair<int,string> PII; typedef long long ll; const int N=20+5,M=998244353; string a[3],b[3]; string ans[3][10]={{" _ "," "," _ "," _ "," "," _ "," _ "," _ "," _ "," _ "}, {"| |"," |"," _|"," _|","|_|","|_ ","|_ "," |","|_|","|_|"}, {"|_|"," |","|_ "," _|"," |"," _|","|_|"," |","|_|"," _|"} }; string P(int x,string s[]){ string y1=s[0].substr(x,3),y2=s[1].substr(x,3),y3=s[2].substr(x,3); if(y1==" _ "&&y2=="| |"&&y3=="|_|")return "0"; if(y1==" "&&y2==" |"&&y3==" |")return "1"; if(y1==" _ "&&y2==" _|"&&y3=="|_ ")return "2"; if(y1==" _ "&&y2==" _|"&&y3==" _|")return "3"; if(y1==" "&&y2=="|_|"&&y3==" |")return "4"; if(y1==" _ "&&y2=="|_ "&&y3==" _|")return "5"; if(y1==" _ "&&y2=="|_ "&&y3=="|_|")return "6"; if(y1==" _ "&&y2==" |"&&y3==" |")return "7"; if(y1==" _ "&&y2=="|_|"&&y3=="|_|")return "8"; if(y1==" _ "&&y2=="|_|"&&y3==" _|")return "9"; } int main(){ cin.tie(0),cout.tie(0); for(int i=0;i<3;++i) getline(cin,a[i]); for(int i=0;i<3;++i) getline(cin,b[i]); string c,d; for(int i=0;i+3<=18;i+=3){ c+=P(i,a); d+=P(i,b); } int a1,b1,c1,a2,b2,c2,t1,t2,t3; a1=(c[0]-'0')*10+c[1]-'0',b1=(c[2]-'0')*10+c[3]-'0',c1=(c[4]-'0')*10+c[5]-'0'; a2=(d[0]-'0')*10+d[1]-'0',b2=(d[2]-'0')*10+d[3]-'0',c2=(d[4]-'0')*10+d[5]-'0'; int q=0; if(a1==a2){ if(b1==b2){ if(c1==c2)cout<<"gang gang hao\n"; else if(c1>c2)cout<<"early\n",q=1; else cout<<"late\n",q=2; } else if(b1>b2)cout<<"early\n",q=1; else cout<<"late\n",q=2; } else if(a1>a2)cout<<"early\n",q=1; else cout<<"late\n",q=2; if(q==1){ t3=c1-c2; if(t3<0){ t3+=60; b1--; } t2=b1-b2; if(t2<0){ t2+=60; a1--; } t1=a1-a2; } else if(q==2){ t3=c2-c1; if(t3<0){ t3+=60; b2--; } t2=b2-b1; if(t2<0){ t2+=60; a2--; } t1=a2-a1; } else t1=t2=t3=0; for(int i=0;i<3;++i){ cout<<ans[i][t1/10]<<ans[i][t1%10]; cout<<ans[i][t2/10]<<ans[i][t2%10]; cout<<ans[i][t3/10]<<ans[i][t3%10]; cout<<'\n'; } return 0; }
简单的 LRU 问题
思路：模拟
#include<bits/stdc++.h> using namespace std; const int N=998244353; typedef long long ll; int n,m,mx; int a[105][10]; string hx = "0123456789abcdef"; int main(){ cin >> n >> m; mx = 0; cout << "+------+"; for (int i = 0; i < n; ++i) { cout << "--------+"; } cout << '\n'; cout << "| |"; for (int i = 0; i < n; ++i) { cout << " 0x"; int a = i / 16, b = i % 16; cout << hx.substr(a, 1) << hx.substr(b, 1) << " |"; } cout << '\n'; cout << "+------+"; for (int i = 0; i < n; ++i) { cout << "--------+"; } cout << '\n'; for (int i = 0; i < m; ++i) { int x; cin >> x; if (i == 0) a[0][0] = x; else { int have = 0; for (int j = 0; j <= mx; ++j) { if (a[i - 1][j] == x) { have = 1; break; } } if (have) { int k = 0; for (int j = 0; j <= mx; ++j) { if (a[i - 1][j] == x) continue; a[i][k++] = a[i - 1][j]; } a[i][mx] = x; } else { if (mx < n - 1) { for (int j = 0; j <= mx; ++j) { a[i][j] = a[i - 1][j]; } a[i][++mx] = x; } else { int k = 0; for (int j = 1; j <= mx; ++j) { a[i][k++] = a[i - 1][j]; } a[i][mx] = x; } } } cout << "| 0x"; int A, B, C, D; A = i / 16; B = i % 16; cout << hx.substr(A, 1) << hx.substr(B, 1) << " |"; for (int j = 0; j <= mx; ++j) { cout << " 0x"; int tmp = a[i][j]; A = tmp / 4096; tmp -= A * 4096; B = tmp / 256; tmp -= B * 256; C = tmp / 16; tmp -= C * 16; D = tmp; cout << hx.substr(A, 1) << hx.substr(B, 1); cout << hx.substr(C, 1) << hx.substr(D, 1) << " |"; } for (int j = mx + 1; j < n; ++j) { cout << " |"; } cout << '\n'; cout << "+------+"; for (int i = 0; i < n; ++i) { cout << "--------+"; } cout << '\n'; } return 0; }
思路：排列喜爱度后输出k+1个
#include<bits/stdc++.h> using namespace std; const int N=1e5+5; typedef pair<int,string> P; vector<P>ve; bool cmp(P a,P b){ return a.first>b.first; } int main(){ int n,k; cin>>n; int x; string a; for(int i=0;i<n;++i){ cin>>x>>a; ve.push_back({x,a}); } sort(ve.begin(),ve.end(),cmp); cin>>k; cout<<ve[k].second; return 0; }
毁灭凤凰人
思路：统计拥有卡牌，根据表示形式比较是否能将毁灭凤凰人送入墓地和除外
#include<bits/stdc++.h> using namespace std; const int N=1e5+5; int n,m; int main(){ cin>>n>>m; int x,y,yy=0,a=0,b=0; for(int i=1;i<=n;++i){ cin>>x; if(x==0){ cin>>y;yy=max(yy,y); } else if(x==1)a=1; else b=1; } if(a&&m==0&&yy>=2500)cout<<"haoye"; else if(a&&m==1&&yy>2100)cout<<"haoye"; else if(b&&n>1)cout<<"haoye"; else cout<<"QAQ"; return 0; }
#include<bits/stdc++.h> using namespace std; const int N=2e5+5; int main(){ int n; cin>>n; for(int i=0;i<n;++i)cout<<"Welcome to HZNU\n"; return 0; }
Ayanoto 变形记
思路：只有x为0时跳不到
#include<bits/stdc++.h> using namespace std; const int N=1e5+5; int main(){ int t,n,x; cin>>t; while(t--){ cin>>n>>x; if(x==0)cout<<"no"<<'\n'; else{ cout<<"yes"<<'\n'; } } return 0; }
P 龙学长的教诲
思路：区分出符号，前后依次输出
#include<bits/stdc++.h> using namespace std; const int N=1e5+5; int t; string a[1005]; int main(){ cin>>t; while(t--){ int n; string x; for(int i=0;;++i){ cin>>a[i]; char b=a[i][a[i].size()-1]; if(b=='.'||b=='!'||b=='?'){ x=b;n=i+1;a[i]=a[i].substr(0,a[i].size()-1); break; } } int z=0,y=n-1; for(int i=1;i<=n;++i){ if(i%2)cout<<a[z++]; else cout<<a[y--]; if(i<n)cout<<' '; }cout<<x<<'\n'; } return 0; }
