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
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"<
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<
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<
区间特性直接双指针
void solve(){
cin>>n>>m;
vector a(n);
for(auto&v:a) cin>>v;
sort(a.begin(),a.end());
int ans=n;
for(int i=0;i
注意二分的细节
void solve(){
cin>>n>>m;
vector 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>1;
if(check(mid)) l=mid;
else r=mid-1;
}
cout<
埃斯筛原理
void solve(){
cin>>n;
map mp;
for(int i=1;i<=n;i++){
int x; cin>>x;
mp[x]++;
}
vector 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<
八皇后对角线的判断
void solve(){
map 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<
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")<
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 st(n+1);
vector>> g(n+1);
vector 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 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")<