Codeforces Round 886 (Div. 4)

目录

A. To My Critics

B. Ten Words of Wisdom

C. Word on the Paper

D. Balanced Round

E. Cardboard for Pictures

F. We Were Both Children

G. The Morning Star

H. The Third Letter

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A. To My Critics

直接模拟

void solve(){int a,b,c; cin>>a>>b>>c;if(a+b>=10||a+c>=10||b+c>=10) cout<<"YES"<<endl;else cout<<"NO"<<endl;return ;
}

B. Ten Words of Wisdom

可以通过顺序遍历直接找到最小值或者是直接按照某个要求排序

struct code{int x,id;bool operator<(const code&t)const{return x>t.x;}
}e[N];void solve(){cin>>n;for(int i=1;i<=n;i++){int a,b; cin>>a>>b;if(a>10) b=-1;e[i]={b,i};}sort(e+1,e+1+n);cout<<e[1].id<<endl;
}

C. Word on the Paper

直接找到目标即可

void solve(){n=8;for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)cin>>s[i][j];for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)if(s[i][j]!='.'){while(i<=n && s[i][j]!='.'){cout<<s[i][j];i++;}cout<<endl;return ;}   }

D. Balanced Round

区间特性直接双指针

void solve(){cin>>n>>m;vector<int> a(n);for(auto&v:a) cin>>v;sort(a.begin(),a.end());int ans=n;for(int i=0;i<n;i++){int j=i;while(j+1<n && a[j+1]-a[j]<=m) j++;ans=min(ans,n-(j-i+1));i=j;}cout<<ans<<endl;}

E. Cardboard for Pictures

注意二分的细节

void solve(){cin>>n>>m;vector<int> a(n);for(auto&v:a) cin>>v;auto check = [&](LL mid){LL ans=0;for(auto&v:a){ans+=(LL)(v+2*mid)*(v+2*mid);if(ans>m) return false;// 防止爆LL直接使用每次都特殊判断}return ans<=m;};LL l=1,r=1e9;while(l<r){LL mid=l+r+1>>1;if(check(mid)) l=mid;else r=mid-1;}cout<<l<<endl;
}

F. We Were Both Children

埃斯筛原理

void solve(){cin>>n;map<int,int> mp;for(int i=1;i<=n;i++){int x; cin>>x;mp[x]++;}vector<int> ton(n+1);for(auto&[v,w]:mp)for(int j=v;j<=n;j+=v)ton[j]+=w;int ans=0;for(auto&v:ton) ans=max(ans,v);cout<<ans<<endl;
}

G. The Morning Star

八皇后对角线的判断

void solve(){map<int,int> mp[5];cin>>n;LL ans=0;while(n--){int x,y; cin>>x>>y;ans+=mp[1][x]+mp[2][y]+mp[3][x+y]+mp[4][x-y];mp[1][x]++;mp[2][y]++;mp[3][x+y]++;mp[4][x-y]++;}ans*=2;cout<<ans<<endl;}

H. The Third Letter

1.注意带权并查集的原理和思维即可

int find(int x){if(x!=p[x]){int t=find(p[x]);d[x]+=d[p[x]];p[x]=t;}return p[x];
}
void solve(){bool ok=true;cin>>n>>m;for(int i=1;i<=n;i++) p[i]=i,d[i]=0;while(m--){int a,b,c; cin>>a>>b>>c;int fa=find(a),fb=find(b);if(fa!=fb){p[fb]=fa;d[fb]=d[a]+c-d[b];}else{if(d[b]-d[a]!=c) ok=false;}}cout<<(ok ? "YES" : "NO")<<endl;
}

2.或者直接使用dfs去找是否有不满足的即可

void solve(){bool ok=true;cin>>n>>m;vector<bool> st(n+1);vector<vector<pair<int,int>>> g(n+1);vector<LL> d(n+1);while(m--){int a,b,c; cin>>a>>b>>c;g[a].push_back({b,c});g[b].push_back({a,-c});}function<void(int u)> dfs = [&](int u){st[u]=true;for(auto&[v,w]:g[u]){if(st[v]){if(d[v]!=d[u]+w){ok=false;return ;}}else{d[v]=d[u]+w;dfs(v);}}};for(int i=1;i<=n;i++)if(!st[i]){dfs(i);}cout<<(ok ? "YES" : "NO")<<endl;
}