Problem A??小A的簽到題

https://ac.nowcoder.com/acm/contest/549/A

C++版本一

題解：

?

C++版本二

題解：矩陣快速冪+斐波那契數列

參考文章：矩陣快速冪 斐波那契數列 斐波那契數列矩陣快速冪法

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int #define endl "\n" using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100000+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; ll t,n,m,k,p,l,r,u,v; int ans,cnt,flag,temp,sum; char str; ll b[2][2]; ll a[2][2]; void Matrix(ll a[2][2],ll b[2][2]){ll c[2][2];memset(c,0,sizeof(c));for(int i=0;i<2;i++){for(int j=0;j<2;j++){for(int k=0;k<2;k++){c[i][j]=c[i][j]+a[i][k]*b[k][j];}}}for(int i=0;i<2;i++){for(int j=0;j<2;j++){a[i][j]=c[i][j];}} } void power(ll k){for(int i=0;i<2;i++){for(int j=0;j<2;j++){a[i][j]=1;b[i][j]=1;}}a[0][0]=b[0][0]=0;while(k){if(k&1)Matrix(a,b);Matrix(b,b);k>>=1;} } int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){scanf("%lld",&n);power(n-1);printf("%lld\n",a[0][1]*a[0][1]-a[0][0]*a[0][0]-a[0][0]*a[0][1]);//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

Problem B?小A的回文串

https://ac.nowcoder.com/acm/contest/549/B

題解：字符串 樸素 回文字符串

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int #define endl "\n" using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100000+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,p,l,r,u,v; int ans,cnt,flag,temp,sum; int a[N]; char str[N]; struct node{}; int check(int n,int pos){int len1=1;int x=pos-1,y=pos+1;while(x>=0&&y<2*n&&len1+2<=n&&str[x]==str[y]){x--;y++;len1+=2;}int len2=0;x=pos,y=pos+1;while(x>=0&&y<2*n&&len2+2<=n&&str[x]==str[y]){x--;y++;len2+=2;}return max(len1,len2); } int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){//scanf("%d",&n);cin>>str;int len=strlen(str);for(int i=0;i<len;i++){str[i+len]=str[i];ans=max(ans,check(len,i));}for(int i=len;i<2*len;i++){ans=max(ans,check(len,i));}cout<<ans<<endl;//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

Problem C?小A買彩票

https://ac.nowcoder.com/acm/contest/549/C

題解：背包問題 分數

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int #define endl "\n" using namespace std; typedef long long ll; //typedef __int128 lll; const int N=1000+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,p,l,r,u,v; ll ans,cnt,flag,temp,sum; ll dp[N][N]; char str; ll gcd(ll a,ll b){//cout<<a<<" "<<b<<endl;if(b==0)return a;return gcd(b,a%b); } struct node{}; int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){dp[1][1]=dp[1][2]=dp[1][3]=dp[1][4]=1;for(int i=2;i<=30;i++){for(int k=1;k<=4;k++){for(int j=i*4;j>=max(i,k);j--){dp[i][j]=dp[i][j]+dp[i-1][j-k];}}}while(~scanf("%d",&n)){if(n==0){cout<<"1/1"<<endl;continue;}sum=1;for(int i=1;i<=n;i++)sum*=4;ans=0;for(int i=3*n;i<=4*n;i++)ans+=dp[n][i];ll GCD=gcd(sum,ans);cout<<ans/GCD<<"/"<<sum/GCD<<endl;}//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

Problem D?小A的位運算

https://ac.nowcoder.com/acm/contest/549/D

題解：預處理了一下前綴和后綴，然后枚舉那個不選的數就可以了。

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int #define endl "\n" using namespace std; typedef long long ll; //typedef __int128 lll; const int N=5000000+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,p,l,r,u,v; int ans,cnt,flag,temp,sum; int a[N]; char str; struct node{}; int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){scanf("%d",&n);for(int i=1;i<=n;i++){scanf("%d",&a[i]);sum|=a[i];}for(int i=n;i>=1;i--){sum|=a[i];if(ans<(sum|temp))ans=sum|temp;temp|=a[i];}cout<<ans<<endl;//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

Problem E?小A的路徑

https://ac.nowcoder.com/acm/contest/549/E

題解：矩陣快速冪+最短路（Floyd算法）

參考文章：矩陣快速冪?Floyd算法?

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int #define endl "\n" using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,p,l,r,u,v; ll ans,cnt,flag,temp,sum; char str; ll b[N][N]; ll a[N][N]; void Matrix(ll a[N][N],ll b[N][N]){ll c[N][N];memset(c,0,sizeof(c));for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){for(int k=1;k<=n;k++){c[i][j]=(c[i][j]+a[i][k]*b[k][j])%MOD;}}}for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){a[i][j]=c[i][j];}} } void power(ll k){for(int i=1;i<=n;i++){a[i][i]=1;}while(k){if(k&1)Matrix(a,b);Matrix(b,b);k>>=1;} } int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){scanf("%d%d%d%d",&n,&m,&k,&t);for(int i=1;i<=m;i++){scanf("%d%d",&u,&v);b[u][v]++;}power(k);for(int i=1;i<=n;i++)if(t!=i)ans=(ans+a[t][i])%MOD;cout<<ans<<endl;//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

Problem F?小A的最短路

https://ac.nowcoder.com/acm/contest/549/F

C++版本一

題解：最近公共祖先+最短路（SPFA算法）

#pragma GCC optimize(3,"Ofast","inline") #include<bits/stdc++.h> #define pb push_back #define Rep(i,a,b) for(int i=a;i<=b;i++) using namespace std; typedef long long ll; typedef long double ld; const int inf=0x3f3f3f3f; const ll INF=9e18; const int N=3e5+50; struct edge{ int v,w; }; vector<edge> G[N]; int n,U,V,dep[N],Fa[N][22];void dfs(int u,int fa) {dep[u]=dep[fa]+1;Fa[u][0]=fa;for(int i=1;(1<<i)<=dep[u];i++) Fa[u][i]=Fa[Fa[u][i-1]][i-1];for(auto &e:G[u]) {int v=e.v;if(v==fa) continue;dfs(v,u);} } int lca(int x,int y) {if(dep[x]<dep[y]) swap(x,y);for(int i=20;i>=0;i--) if((1<<i)<=dep[x]-dep[y]) x=Fa[x][i];if(x==y) return x;for(int i=20;i>=0;i--) if(Fa[x][i]!=Fa[y][i]) x=Fa[x][i],y=Fa[y][i];return Fa[x][0]; } int solve(int x,int y) { return dep[x]+dep[y]-2*dep[lca(x,y)]; }struct node{int u,dis;node() {}node(int u,int dis):u(u),dis(dis) {}bool operator <(const node &rhs) const {return dis>rhs.dis;} }; int dis[N]; priority_queue<node> q; void spfa() {for(int i=1;i<=n;i++) dis[i]=inf;dis[U]=0;q.push({U,0});while(!q.empty()) {int u=q.top().u,d=q.top().dis;q.pop();if(dis[u]<d) continue;for(auto &e:G[u]) {int v=e.v,w=e.w;if(dis[v]>dis[u]+w) {dis[v]=dis[u]+w;q.push({v,dis[v]});}}} } int main() {scanf("%d",&n);for(int i=1;i<n;i++) {int u,v;scanf("%d%d",&u,&v);G[u].push_back({v,1});G[v].push_back({u,1});}dfs(1,0);scanf("%d%d",&U,&V);G[U].push_back({V,0});G[V].push_back({U,0});spfa();int q;scanf("%d",&q);while(q--) {int x,y;scanf("%d%d",&x,&y);printf("%d\n",min(solve(x,y),dis[x]+dis[y]));}return 0; }

C++版本二?

題解：最近公共祖先+最短路（dijkstra算法）

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int #define endl "\n" using namespace std; typedef long long ll; //typedef __int128 lll; const int N=300000+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,p,l,r,u,v; int ans,cnt,flag,temp,sum; int dep[N]; int dp[N][23]; int dis[N]; bool vis[N]; bool VIS[N]; char str; struct node{int e,w;bool operator <(const node &S)const{return w>S.w;} }x,tmp; vector<node>G[N]; void dfs(int u){vis[u]=1;for(int i=0,j=G[u].size();i<j;i++){int v=G[u][i].e;if(vis[v])continue;dep[v]=dep[u]+1;dp[v][0]=u;dfs(v);} } void RMQ(){for(int i=0;i<22;i++){for(int j=1;j<=n;j++){dp[j][i+1]=dp[dp[j][i]][i];}} } int LCA(int x,int y){if(dep[x]<dep[y])swap(x,y);//while(dep[x]>dep[y])x=dp[x][(int)log2(dep[x]-dep[y])];if(x==y)return x;for(int i=log2(dep[x]);i>=0;i--){if(dp[x][i]!=dp[y][i])x=dp[x][i],y=dp[y][i];}return dp[x][0]; } void init(){for(int i=1;i<=n;i++){G[i].clear();}memset(vis,0,sizeof(vis));memset(dep,0,sizeof(dep));memset(dp,0,sizeof(dp));memset(dis,0,sizeof(dis));memset(VIS,0,sizeof(VIS)); } int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){scanf("%d",&n);init();for(int i=1;i<n;i++){scanf("%d%d",&u,&v);x.e=v;x.w=1;G[u].push_back(x);x.e=u;G[v].push_back(x);}dfs(1);scanf("%d%d",&u,&v);RMQ();priority_queue<node>q;x.e=u;x.w=0;q.push(x);x.e=v;q.push(x);while(!q.empty()){x=q.top();q.pop();if(VIS[x.e])continue;VIS[x.e]=1;dis[x.e]=x.w;for(int i=0,j=G[x.e].size();i<j;i++){tmp=G[x.e][i];if(VIS[tmp.e])continue;tmp.w=x.w+tmp.w;q.push(tmp);}}scanf("%d",&m);for(int i=1;i<=m;i++){scanf("%d%d",&u,&v);cout<<min(dep[u]+dep[v]-2*dep[LCA(u,v)],dis[u]+dis[v])<<endl;}//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

Problem G?小A與小B

https://ac.nowcoder.com/acm/contest/549/G

題解：DFS

1、時間不一定相同；

2、小B最后相遇時刻可以走一步；

3、小B兩步都不能在障礙物上；

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int using namespace std; typedef long long ll; //typedef __int128 lll; const int N=2000+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,q; int ans,cnt,flag,temp,sum; int a[N],b[N]; bool vis1[N][N],vis2[N][N]; int dis[N][N]; char str[N][N]; int ax,ay,bx,by; struct node{int x,y,time; }s,f,tmp; int dir2[4][2]={1,0,0,1,-1,0,0,-1}; int dir1[8][2]={1,0,0,1,-1,0,0,-1,1,1,-1,-1,1,-1,-1,1};int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){scanf("%d%d",&n,&m);for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){cin>>str[i][j];if(str[i][j]=='C'){ax=i;ay=j;}if(str[i][j]=='D'){bx=i;by=j;}}}queue<node>q;s.x=ax;s.y=ay;s.time=0;q.push(s);vis1[ax][ay]=1;while(!q.empty()){f=q.front();q.pop();for(int i=0;i<8;i++){tmp.x=f.x+dir1[i][0];tmp.y=f.y+dir1[i][1];tmp.time=f.time+1;if(tmp.x<1||tmp.x>n||tmp.y<1||tmp.y>m)continue;if(vis1[tmp.x][tmp.y])continue;if(str[tmp.x][tmp.y]=='#')continue;vis1[tmp.x][tmp.y]=1;dis[tmp.x][tmp.y]=tmp.time;q.push(tmp);}}//for(int i=1;i<=n;i++){//for(int j=1;j<=m;j++){//printf("%d%c",dis[i][j]," \n"[j==m]);//}//}while(!q.empty())q.empty();s.x=bx;s.y=by;s.time=0;q.push(s);vis2[bx][by]=1;ans=INF;while(!q.empty()){f=q.front();q.pop();//cout<<f.x<<" "<<f.y<<endl;if(vis1[f.x][f.y]){ans=min(max(dis[f.x][f.y],f.time),ans);flag=1;//cout<<ans<<endl;}for(int i=0;i<4;i++){tmp.x=f.x+dir2[i][0];tmp.y=f.y+dir2[i][1];tmp.time=f.time+1;if(tmp.x<1||tmp.x>n||tmp.y<1||tmp.y>m)continue;if(vis2[tmp.x][tmp.y])continue;if(str[tmp.x][tmp.y]=='#')continue;if(vis1[tmp.x][tmp.y]){ans=min(max(dis[tmp.x][tmp.y],tmp.time),ans);flag=1;//cout<<ans<<endl;}for(int j=0;j<4;j++){tmp.x=f.x+dir2[i][0]+dir2[j][0];tmp.y=f.y+dir2[i][1]+dir2[j][1];if(tmp.x<1||tmp.x>n||tmp.y<1||tmp.y>m)continue;if(vis2[tmp.x][tmp.y])continue;if(str[tmp.x][tmp.y]=='#')continue;if(vis1[tmp.x][tmp.y]){ans=min(max(dis[tmp.x][tmp.y],tmp.time),ans);flag=1;//cout<<ans<<endl;}vis2[tmp.x][tmp.y]=1;q.push(tmp);}}}if(flag){cout<<"YES"<<endl;cout<<ans<<endl;}else{cout<<"NO"<<endl;}//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

?

Problem H?小A的柱狀圖

https://ac.nowcoder.com/acm/contest/549/H

題解：單調棧

?

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int #define endl "\n" using namespace std; typedef long long ll; //typedef __int128 lll; const int N=1000000+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,p,l,r,u,v; ll ans,cnt,flag,temp,sum; int a[N]; int b[N]; int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){scanf("%d",&n);for(int i=1;i<=n;i++){scanf("%d",&a[i]);}for(int i=1;i<=n;i++){scanf("%d",&b[i]);}stack<int>s,t;s.push(0);t.push(0);for(int i=1;i<=n;i++){ll w=0;while(s.top()>b[i]){w+=t.top();ans=max(ans,(ll)w*s.top());s.pop();t.pop();}s.push(b[i]);t.push(w+a[i]);}cout<<ans<<endl;//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

Problem I?小A取石子

https://ac.nowcoder.com/acm/contest/549/I

題解：NIM游戲 博弈論

判斷一下小A是否能夠取石子，當且僅當小A原本就是必敗態并且不能取出任何石子的情形下，小A會輸否則小A都會贏。

參考文章：SG函數? ? NIM游戲

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int #define endl "\n" using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100000+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,p,l,r,u,v; int ans,cnt,flag,temp,sum; int a[N]; char str; struct node{}; int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){scanf("%d%d",&n,&k);for(int i=1;i<=n;i++){scanf("%d",&a[i]);sum^=a[i];}if(sum)ans=1;for(int i=1;i<=n;i++){if(a[i]>=k){temp=sum^a[i]^(a[i]-k);if(temp){ans=1;break;}}}cout<<(ans?"YES":"NO")<<endl;//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

Problem J?小A的數學題

https://ac.nowcoder.com/acm/contest/549/J

題解：

#pragma GCC optimize(3,"Ofast","inline") #include<bits/stdc++.h> #define pb push_back #define Rep(i,a,b) for(int i=a;i<=b;i++) using namespace std; typedef long long ll; typedef long double ld; const int inf=0x3f3f3f3f; const ll INF=9e18; const int N=1e6+50; const int maxn=1e6+50; const ll mod=1e9+7; bool vis[maxn]; int cnt=0,pri[maxn],mu[maxn]; inline void init(){mu[1]=1;for(int i=2;i<=(int)1e6+1;i++){if(!vis[i]) pri[++cnt]=i,mu[i]=-1;for(int j=1;j<=cnt&&pri[j]*i<=(int)1e6+1;j++){vis[i*pri[j]]=true;if(i%pri[j]==0){ mu[i*pri[j]]=0;break; }else mu[i*pri[j]]=-mu[i];}} } int main() {init();ll n,m,ans=0;scanf("%lld%lld",&n,&m);ll k=min(m,n);for(ll i=1;i<=k;i++) {for(ll j=i;j<=k;j+=i) {ans+=i*i*(ll)mu[j/i]*(n/j)*(m/j);ans%=mod;}}printf("%lld\n",ans);return 0; }

?

總結

以上是生活随笔為你收集整理的牛客小白月赛13的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。