2023第十四届蓝桥杯C++B组题目解析

起因

24年蓝桥杯将至，把去年题复习一遍。emmm，其实是有一个很厉害的学妹说“要被圈钱杯把钱圈光了”。
现在只是初稿，如果有需要细写解析的可以在下面评论，欢迎留言。

日期统计

#include<bits/stdc++.h>
typedef  long long LL ;
const int N = 1e5 + 10 , INF = 1e9 + 7 ;
using namespace  std ;
int  n , m ;int date[ 618 ] = {5, 6, 8, 6, 9, 1, 6, 1, 2, 4, 9, 1, 9, 8, 2, 3, 6, 4, 7, 7,5, 9, 5, 0, 3, 8, 7, 5, 8, 1, 5, 8, 6, 1, 8, 3, 0, 3, 7, 9,2, 7, 0, 5, 8, 8, 5, 7, 0, 9, 9, 1, 9, 4, 4, 6, 8, 6, 3, 3,8, 5, 1, 6, 3, 4, 6, 7, 0, 7, 8, 2, 7, 6, 8, 9, 5, 6, 5, 6,1, 4, 0, 1, 0, 0, 9, 4, 8, 0, 9, 1, 2, 8, 5, 0, 2, 5, 3, 3
};
int monthNum[ 13 ] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31
};void AC(){LL ans = 0;for (int month = 1 ; month <= 12 ; ++month) {for (int day = 1 ; day <= monthNum[ month ] ; ++day ) {int realdate[ 8 ] = {2, 0, 2, 3, month / 10, month % 10, day / 10, day % 10};int k = 0;for (int i = 0; i < 100; ++i) {if ( date[ i ] == realdate[ k ] ) {++k;if ( k == 8 ) {ans++;break;}}}}}cout << ans << endl;
}int main(){ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int t  = 1 ;//cin >> t ;while( t-- ){AC();}return 0 ;
}

01串的熵

#include<bits/stdc++.h>
typedef  long long LL ;
const int N = 1e5 + 10 , INF = 1e9 + 7 ;
const double ans = 11625907.5798 , eps = 1e-2 ;
using namespace  std ;
int  n , m ;void AC(){n = 23333333 ;for( int i = 0 ; i <= n / 2 ; i ++ ){ // i 对应0的个数double nn = n ; // 强转n的类型double sum = i * ( i / nn ) * log2( i / nn ) + ( nn - i )*( ( nn - i ) / nn ) * log2( ( nn - i ) / nn );sum = -sum ;//cout << "su m= " << sum << endl;if( abs( sum - ans ) <= eps ){cout << i << endl;return ;}}
}int main(){ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int t  = 1 ;//cin >> t ;while( t-- ){AC();}return 0 ;
}

冶炼金属

#include<bits/stdc++.h>
typedef  long long LL ;
const int N = 1e5 + 10 , INF = 1e9 + 7 ;
const double ans = 11625907.5798 , eps = 1e-2 ;
using namespace  std ;
int  n , m ;void AC(){cin >> n ;int mx = INF , mn = 1 ;for( int i = 1 ; i <= n ; i++ ){int x , y ; cin >> x >> y ;mx = min( mx , x / y ) ;mn = max( mn , ( x / ( y + 1 ) ) + 1 ) ;}cout << mn << " " << mx << endl;
}int main(){ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int t  = 1 ;//cin >> t ;while( t-- ){AC();}return 0 ;
}

飞机降落

#include<bits/stdc++.h>
typedef  long long LL ;
const int N = 1e5 + 10 , INF = 1e9 + 7 ;
const double ans = 11625907.5798 , eps = 1e-2 ;
using namespace  std ;
int  n , m ;
struct op{int st , ed , cont ;
}f[ 15 ];
bool vis[ 15 ] , fg ;void dfs( int cnt , int now_time ){if( cnt == n || fg ){fg = true ;return ;}for( int i = 1 ; i <= n ; i++ ){if( !vis[ i ] && now_time <= f[ i ].ed ){vis[ i ] = true ;dfs( cnt + 1 , max( now_time , f[ i ].st  ) + f[ i ].cont ) ;vis[ i ] = false ;}}
}void AC(){cin >> n ;for( int i = 1 ; i <= n ; i++ ){cin >> f[ i ].st >> f[ i ].ed >> f[ i ].cont ;f[ i ].ed += f[ i ].st ;}fg = false ;dfs(  0 , 0 ) ;if( fg ){cout << "YES\n";}else{cout << "NO\n";}
}int main(){ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int t  = 1 ;cin >> t ;while( t-- ){AC();}return 0 ;
}

接龙序列

#include<bits/stdc++.h>
typedef  long long LL ;
const int N = 1e5 + 10 , INF = 1e9 + 7 ;
const double ans = 11625907.5798 , eps = 1e-2 ;
using namespace  std ;
int  n , m ;
int dp[ 15 ] ;
void AC(){cin >> n ;int ans = 0 ; string s ;for( int i = 1 ; i <= n ;i ++ ){cin >> s ;int pre = s[ 0 ] , hou = s[ (int)s.size() - 1 ] ;dp[ hou ] = max( dp[ hou ] , dp[ pre ] + 1 ) ;ans = max( ans , dp[ hou ] ) ;}cout << n - ans << endl ;return ;
}int main(){ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int t  = 1 ;//cin >> t ;while( t-- ){AC();}return 0 ;
}

岛屿个数

这题不想写，暂时套用别人代码（）。

#include <bits/stdc++.h>
#define MAX 51
using namespace std;
int T,m,n;
int vis[MAX][MAX],used[MAX][MAX];
int dx[]={1,-1,0,0,1,1,-1,-1};
int dy[]={0,0,1,-1,1,-1,1,-1};
string mp[MAX];
void bfs_col(int x,int y){queue<int>qx,qy;qx.push(x);qy.push(y);vis[x][y]=1;while(!qx.empty()){x=qx.front();qx.pop();y=qy.front();qy.pop();for(int i=0;i<4;i++){int nx=x+dx[i];int ny=y+dy[i];if(nx<0||ny<0||nx>m-1||ny>n-1||vis[nx][ny]==1||mp[x][y]=='0') continue;qx.push(nx);qy.push(ny);vis[nx][ny]=1;}}
}
bool bfs_out(int x,int y){for(int i=0;i<m;i++){for(int j=0;j<n;j++){used[i][j]=0;}}queue<int>qx,qy;qx.push(x);qy.push(y);used[x][y]=1;while(!qx.empty()){x=qx.front();qx.pop();y=qy.front();qy.pop();if(x==0||x==m-1||y==0||y==n-1) return true;for(int i=0;i<8;i++){int nx=x+dx[i];int ny=y+dy[i];if(nx<0||ny<0||nx>m-1||ny>n-1||used[nx][ny]==1||mp[nx][ny]=='1') continue;qx.push(nx);qy.push(ny);used[nx][ny]=1;}}return false;
}
void solve(){int ans=0;cin>>m>>n;for(int i=0;i<m;i++){cin>>mp[i];  for(int j=0;j<n;j++){vis[i][j]=0;}}for(int i=0;i<m;i++){for(int j=0;j<n;j++){if(!vis[i][j]&&mp[i][j]=='1'){bfs_col(i,j);if(bfs_out(i,j)) ans++;}}}cout<<ans<<'\n';
}
int main(){ios::sync_with_stdio(0);cin.tie(0);int T;cin>>T;while(T--)    solve();return 0;
}

子串简写

#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 5e5+10;
int k;
int cntb[N]; 
string a;
char x,y;int main()
{int cnta = 0;cin >> k;cin >> a >> x >> y;cntb[a.size()] = 0;for(int i = a.size()-1; i >= 0; i--){cntb[i] = cntb[i+1];        //cntb[i]来统计包括下标i及之后给出字母y的个数if(a[i] == y){cntb[i] ++;}}long long ans = 0;for(int i = 0; i < a.size(); i++){if(a[i] == x && i+k-1 <= a.size()-1 )   //如果这个a[i]==x  那就看它i+k-1及后面有多少个y，及是cntb[i+k-1]。{ans += cntb[i+k-1];}                //还可以统计一下x的个数，如果没有x了，就可以直接跳出循环。}cout << ans <<endl;    //输出结果return 0;
}

整数删除

#include<bits/stdc++.h>
typedef  long long LL ;
const int N = 5e5 + 10 , INF = 1e9 + 7 ;
using namespace  std ;
LL  n , m , k , l[ N ] , r[ N ] , ans[ N ] , ad[ N ] ;
struct op{LL num , pos ;bool operator < ( const op& now )const{if( num == now.num ){return pos > now.pos ;}return num > now.num ;}
};
priority_queue< op > qu;void AC(){cin >> n >> k ;for( int i = 1; i <= n ; i ++ ){int x ; cin >> x ;l[ i ] = i - 1 ; r[ i ] = i + 1 ;qu.push( { x , i } ) ;}while( (int)qu.size() > n - k ){auto [ x , y ] = qu.top() ; qu.pop() ;if( ad[ y ] ){qu.push( { x +ad[ y ] , y } ) ;ad[ y ] = 0 ;}else{LL nl = l[ y ] , nr = r[ y ] ;ad[ nl ] += x ; ad[ nr ] += x ;l[ nr ] = nl , r[ nl ] = nr ;}}while( (int)qu.size() ){auto[ x , y ] = qu.top() ; qu.pop() ;ans[ y ] = x + ad[ y ] ;}for( LL i = 1 ; i <= n ; i++ ){if( ans[ i ] ) cout << ans [ i ] << " " ;}return ;
}int main(){ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int t  = 1 ;//cin >> t ;while( t-- ){AC();}return 0 ;
}

景区导游（挂）

#include <bits/stdc++.h>
using namespace std;
using ll=long long;
const ll N=1e5+9,P=1e9+7,inf=2e18;
ll n,k,u,v,t,a[N],dep[N],dist[N],fa[N][18],sum;
vector<pair<ll,ll>> g[N];void dfs(ll x,ll p,ll d){fa[x][0]=p;dist[x]=d;for(int i=1;i<=17;++i){fa[x][i]=fa[fa[x][i-1]][i-1];}for(int i=0;i<=g[x].size()-1;++i){if(g[x][i].first==p)continue;dep[g[x][i].first]=dep[x]+1;dfs(g[x][i].first,x,d+g[x][i].second);}
}
ll lca(ll u,ll v){if(dep[u]<dep[v])swap(u,v);for(int i=17;i>=0;--i){if(dep[fa[u][i]]>=dep[v]) u=fa[u][i];}for(int i=17;i>=0;--i){if(fa[u][i]!=fa[v][i])u=fa[u][i],v=fa[v][i];}if(u==v)return u;return fa[u][0];
}
ll getd(ll u,ll v){return dist[u]+dist[v]-dist[lca(u,v)]*2;
}
int main(){ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);cin>>n>>k;for(int i=1;i<=n-1;++i){cin>>u>>v>>t;g[u].push_back({v,t});g[v].push_back({u,t});}dfs(1,0,0);for(int i=1;i<=k;++i)cin>>a[i];//for(int i=1;i<=n;++i){//    cout<<dep[i]<<' '<<dist[i]<<'\n';//}for(int i=1;i<=k-1;++i){sum+=getd(a[i],a[i+1]);}//cout<<sum;for(int i=1;i<=k;++i){ll _sum=sum;if(i!=k)_sum-=getd(a[i],a[i+1]);if(i!=1)_sum-=getd(a[i-1],a[i]);if(i!=1&&i!=k)_sum+=getd(a[i-1],a[i+1]);cout<<_sum<<' ';}return 0;
}

砍树（挂）

#include <bits/stdc++.h>
#define x first
#define y second
using namespace std;typedef long long i64;
typedef array<int, 3> arr;
typedef pair<int, int> PII;const int N = 1e5 + 10;int n, m;
vector<PII> e[N];
int l[N], r[N], id[N];
int sz[N], hs[N], tot, top[N], dep[N], fa[N];
int cnt[N], s[N];// 第一遍DFS，子树大小，重儿子，父亲，深度
void dfs1(int u,int f) {sz[u] = 1;hs[u] = -1;fa[u] = f;dep[u] = dep[f] + 1;for (auto [v, id] : e[u]) {if (v == f) continue;dfs1(v, u);sz[u] += sz[v];if (hs[u] == -1 || sz[v] > sz[hs[u]])hs[u] = v;}
}// 第二遍DFS，每个点DFS序，重链上的链头的元素。
void dfs2(int u, int t) {top[u] = t;l[u] = ++tot;id[tot] = u;if (hs[u] != -1) {dfs2(hs[u], t);}for (auto [v, id] : e[u]) {if (v != fa[u] && v != hs[u]) {dfs2(v, v);}}r[u] = tot;
}int LCA(int u, int v) {while (top[u] != top[v]) {if (dep[top[u]] < dep[top[v]]) v = fa[top[v]];else u = fa[top[u]];}if (dep[u] < dep[v]) return u;else return v;
}void dfs_cal(int u, int f) {for (auto [v, id] : e[u]) if (v != f) {dfs_cal(v, u);cnt[id] = s[v];s[u] += s[v];}
}signed main() {cin.tie(nullptr)->ios::sync_with_stdio(false);cin >> n >> m;for (int i = 1; i < n; i++) {int u, v; cin >> u >> v;e[u].push_back({v, i});e[v].push_back({u, i});}dfs1(1, 0);dfs2(1, 1);for (int i = 1; i <= m; i++) {int u, v; cin >> u >> v;int lca = LCA(u, v);s[u]++, s[v]++, s[lca] -= 2;}dfs_cal(1, 0);int ans = -1;for (int i = 1; i <= n - 1; i++) if (cnt[i] == m) {ans = max(i, ans);}cout << ans << '\n';return 0;    
}