2023 CCPC Henan Provincial Collegiate Programming Contest
Posted sakuya726
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因为是我和队友VP的这场比赛，最后记录的也是比赛时候的提交代码，因此会出现多种不同的码风也是正常现象。
A.小水獭游河南
#include<bits/stdc++.h> using namespace std; #define rg register #define maxn 150500 #define mod 998244353 #define pi 3.141592653 inline int read() int x=0,f=1; char c=getchar(); while(c<\'0\'||c>\'9\') if(c==\'-\') f=-1; c=getchar(); while(c>=\'0\'&&c<=\'9\') x=(x<<3)+(x<<1)+c-48; c=getchar(); return x*f; int t; string s; int num[1500]; int tag1,tag2=-1;//大小写的分割点 inline bool check(string a)//判断回文 int len=a.length(); int l=0,r=len-1; while(l<=r) if(a[l]!=a[r]) return false; l++; r--; return true; inline void pre_work() memset(num,0,sizeof(num)); tag1=s.length(); tag2=-1; for(rg int i=0;i<s.length();++i) num[s[i]]++; if(num[s[i]]>=2||s[i]<\'a\'||s[i]>\'z\') tag1=i; break; memset(num,0,sizeof(num)); for(rg int i=s.length()-1;i>=0;--i) num[s[i]]++; if(num[s[i]]>=2||s[i]<\'a\'||s[i]>\'z\') tag2=i; break; int main() t=read(); while(t--) cin>>s; pre_work(); for(rg int i=0;i<s.length()-1;++i) //[0,i]--[i+1,len-1] if(i<tag1&&check(s.substr(i+1,s.length()-(i+1)))==1) cout<<"HE"<<endl; goto ed; else if((i+1)>tag2&&check(s.substr(0,i+1))==1) cout<<"HE"<<endl; goto ed; else if(i>=tag1&&i+1<=tag2) break; cout<<"NaN"<<endl; ed:;
B.Art for Rest
#include<bits/stdc++.h> using namespace std; #define rg register #define maxn 1000500 #define mod 998244353 #define pi 3.141592653 inline int read() int x=0,f=1; char c=getchar(); while(c<\'0\'||c>\'9\') if(c==\'-\') f=-1; c=getchar(); while(c>=\'0\'&&c<=\'9\') x=(x<<3)+(x<<1)+c-48; c=getchar(); return x*f; int n,m,a[maxn]; int f[maxn][22]; int g[maxn][22]; long long lg[maxn]; inline void ST_prework() for(rg int j=1;(1<<j)<=n;++j) for(rg int i=1;i<=n-(1<<j)+1;++i) f[i][j]=max(f[i][j-1],f[i+(1<<(j-1))][j-1]); g[i][j]=min(g[i][j-1],g[i+(1<<(j-1))][j-1]); inline long long query1(int l,int r) long long k=lg[r-l+1]/(lg[2]); return max(f[l][k],f[r-(1<<k)+1][k]); inline long long query2(int l,int r) long long k=lg[r-l+1]/(lg[2]); return min(g[l][k],g[r-(1<<k)+1][k]); bool xd[maxn]; int ans; int main() n=read(); lg[1]=0; lg[2]=1; for(rg int i=3;i<=n+1;++i) lg[i]=lg[i/2]+1; for(rg int i=1;i<=n;++i) a[i]=read(); f[i][0]=a[i]; g[i][0]=a[i]; ST_prework(); for(rg int k=1;k<=n;++k) if(xd[k]==1) ans++; continue; else bool flag=1; int l=min(n,k+1),r=min(n,2*k); int maxx=query1(1,k); int m=ceil(double(1.0*n)/double(1.0*k))-1; while(m--) if(query2(l,r)<maxx) flag=0; break; else maxx=query1(l,r); l=min(n,l+k); r=min(n,r+k); if(flag==1) ++ans; for(rg int j=k;j<=n;j+=k) xd[j]=1; cout<<ans;
E.矩阵游戏
#include <bits/stdc++.h> #include <ext/rope> #include <ext/pb_ds/assoc_container.hpp> using namespace std; using namespace __gnu_cxx; using namespace __gnu_pbds; #define fi first #define se second #define lc u << 1 #define rc u << 1 | 1 // #define int long long // #define double long long // #define int __int128_t typedef long long LL; const int N = 500 + 10; // const int R = 999997; const int Base = N / 2; const int M = 1e6 + 10; // const int P = 1 << 10; const int INF = 2147483647; typedef unsigned long long ULL; const double eps = 1e-4; const double PI = acos(-1); const int mod = 1e9 + 7; int n, k; int d; int m; int Q; int target; // int p = INF; // __int128_t a = 1; // rope<int> r; tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> treap; char g[N][N]; void solve() scanf("%d %d %d", &n, &m, &k); vector<vector<vector<int>>> f(2, vector<vector<int>>(m + 1, vector<int>(k + 1))); for (int i = 1; i <= n; i++) scanf("%s", g[i] + 1); for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) for (int z = 0; z <= k; z++) if (g[i][j] == \'0\') f[i & 1][j][z] = max(f[(i - 1) & 1][j][z], f[i & 1][j - 1][z]); else if (g[i][j] == \'1\') f[i & 1][j][z] = max(f[(i - 1) & 1][j][z], f[i & 1][j - 1][z]) + 1; else if (z >= 1) f[i & 1][j][z] = max(f[(i - 1) & 1][j][z - 1], f[i & 1][j - 1][z - 1]) + 1; else f[i & 1][j][z] = max(f[(i - 1) & 1][j][z], f[i & 1][j - 1][z]); printf("%d\\n", f[n & 1][m][k]); signed main() int t = 1; scanf("%lld", &t); while (t--) solve(); return 0;
F.Art for Last
#include<bits/stdc++.h> using namespace std; //#define int long long typedef long long LL; const int N=5e5+10; int a[N],b[N],q[N]; int n,k; signed main() cin>>n>>k; for(int i=1;i<=n;i++) scanf("%d",&a[i]); sort(a+1,a+n+1); LL res=1e18; int tt=-1,hh=0; for(int i=2;i<=n;i++) b[i]=abs(a[i]-a[i-1]); if(k>2) while(hh<=tt&&i+1-k>q[hh]) hh++;//判断q[hh]有没有滑出窗口 while(hh<=tt&&b[q[tt]]>=b[i]) tt--;//保证窗口内的数从小到大排序 q[++tt]=i; if(i>=k) res=min(res,(LL)b[q[hh]]*abs(a[i]-a[i-k+1])); else res=min(res,(LL)b[i]*b[i]); cout<<res;
H.Travel Begins
#include<bits/stdc++.h> using namespace std; #define rg register #define maxn 250500 #define mod 998244353 #define pi 3.141592653 inline int read() int x=0,f=1; char c=getchar(); while(c<\'0\'||c>\'9\') if(c==\'-\') f=-1; c=getchar(); while(c>=\'0\'&&c<=\'9\') x=(x<<3)+(x<<1)+c-48; c=getchar(); return x*f; int t; int n,k; int minn,maxx; int main() t=read(); while(t--) n=read(); k=read(); minn=maxx=0; int tot1=k-1; if(tot1%2==0) int res=n-tot1/2; cout<<max(0,res)<<" "; else int res=n-(tot1-1)/2; cout<<max(0,res)<<" "; // cout<<k-1<<" "<<(((k-1)%2==0)?(n-(k-1)/2):(n-(k-2)/2))<<endl; cout<<min(2*n,(k-1)+(((k-1)%2==0)?(n-(k-1)/2):(n-(k-2)/2)))<<endl;
K.排列与质数
#include <bits/stdc++.h> #include <ext/rope> #include <ext/pb_ds/assoc_container.hpp> using namespace std; using namespace __gnu_cxx; using namespace __gnu_pbds; #define fi first #define se second #define lc u << 1 #define rc u << 1 | 1 // #define int long long // #define double long long // #define int __int128_t typedef long long LL; const int N = 1e5 + 10; // const int R = 999997; const int Base = N / 2; const int M = 1e6 + 10; // const int P = 1 << 10; const int INF = 2147483647; typedef unsigned long long ULL; const double eps = 1e-4; const double PI = acos(-1); const int mod = 1e9 + 7; int n, k; int d; int m; int Q; int target; // int p = INF; // __int128_t a = 1; // rope<int> r; tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> treap; int primes[N]; int cnt = 0; bool st[N]; bool isp[N]; void init() for (int i = 2; i <= 1e5; i++) if (!st[i]) st[i] = true; primes[cnt++] = i; isp[i] = true; for (int j = 0; primes[j] <= 1e5 / i; j++) st[primes[j] * i] = true; if (i % primes[j] == 0) break; int find(int n) int res = 2 * (n - 2); // cout << res << endl; for (int i = 0; i < cnt; i++) int j = primes[i]; cout << j << endl; if (res + j > n) return 0; if (isp[res] && isp[res + j]) return j; return 0; void solve() scanf("%d", &n); vector<int> a(n + 10); a[1] = 1; if (n <= 4) printf("-1\\n"); return; if (n == 5) printf("4 1 3 5 2\\n"); else if (n == 6) printf("4 6 1 3 5 2\\n"); else if (n == 7) printf("4 6 1 3 5 2 7\\n"); else if (n == 8) printf("4 6 8 1 3 5 2 7\\n"); else if (n == 9) printf("4 6 8 1 3 5 2 7 9\\n"); else if (n == 10) printf("8 6 9 4 1 3 5 2 7 10\\n"); else if (n == 11) printf("8 6 4 1 3 10 5 2 7 9 11\\n"); else if (n == 12) printf("8 11 6 4 1 3 5 2 7 9 12 10\\n"); else if (n & 1) a[4] = 2; a[3] = 5; a[5] = 7; int t = n - 6; int idx = 0; int cur = 0; for (int i = 2; i <= (n + 3) / 2 + 1; i++) if (a[i - 2] == t) a[i] = t + 2; continue; if (!a[i]) a[i] = a[i - 1] + 2; if (a[i] == t) i++; a[i] = n - 1; // cout << i << \' \' << a[i] << endl; cur = n - 1; for (int i = (n + 3) / 2 + 2; i <= n && cur > 0; i++, cur -= 2) a[i] = cur - 2; // for (int i = 1; i <= n; i++) // // cout << i << \' \' << a[i] << endl; // else a[4] = 2; a[3] = 5; a[5] = 7; int t = n - 6; int cur = 4; for (int i = n; i >= n / 2 + 1; i--, cur += 2) if (a[i + 2] == t) a[i] = t + 2; continue; if (!a[i]) a[i] = cur; if (a[i] == t) i--; a[i] = n - 1; // cout << i << \' \' << a[i] << endl; for (int i = 2; i <= n / 2; i++) if (!a[i]) a[i] = a[i - 1] + 2; // cout << i << \' \' << a[i] << endl; for (int i = 1; i <= n; i++) printf("%d", a[i]); if (i != n) printf(" "); else puts(""); signed main() int t = 1; // str = "codeforces"; // init(); // cout << cnt << endl; // init(); // scanf("%d", &t); // getchar(); // int a = 1; // for (int i = 1; i <= 26; i++) // // a = (a << 1) + 1; // // cout << a << endl; // float t = 134217727; // int cnt = 200; // printf("%.16lf", t); // while (cnt--) // // t = t * 2 + a; // printf("%.16f\\n", t); // // getchar(); // cout << (int)(log(4) / log(2)) << endl; while (t--) // cout << t << endl; // // cout << (2563 % 11) << endl; // // cout << t << endl; solve(); // // cout << (((float)1) << 63) << endl; // // cout << \'\\0\' << endl; // // cout << f(4); // // cout<<(25 >> 5) // // cout << gcd(31415, 14142);// // // cout << (28284 / 11) << endl; return 0; /* * __----~~~~~~~~~~~------___ * . . ~~//====...... __--~ ~~ * -. \\_|// |||\\\\ ~~~~~~::::... /~ * ___-==_ _-~o~ \\/ ||| \\\\ _/~~- * __---~~~.==~||\\=_ -_--~/_-~|- |\\\\ \\\\ _/~ * _-~~ .=~ | \\\\-_ \'-~7 /- / || \\ / * .~ .~ | \\\\ -_ / /- / || \\ / * / ____ / | \\\\ ~-_/ /|- _/ .|| \\ / * |~~ ~~|--~~~~--_ \\ ~==-/ | \\~--===~~ .\\ * \' ~-| /| |-~\\~~ __--~~ * |-~~-_/ | | ~\\_ _-~ /\\ * / \\ \\__ \\/~ \\__ * _--~ _/ | .-~~____--~-/ ~~==. * ((->/~ \'.|||\' -_| ~~-/ , . _|| * -_ ~\\ ~~---l__i__i__i--~~_/ * _-~-__ ~) \\--______________--~~ * //.-~~~-~_--~- |-------~~~~~~~~ * //.-~~~--\\ * ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ * * 神兽保佑 永无BUG */ /** * ┏┓ ┏┓+ + * ┏┛┻━━━┛┻┓ + + * ┃ ┃ * ┃ ━ ┃ ++ + + + * ████━████+ * ◥██◤ ◥██◤ + * ┃ ┻ ┃ * ┃ ┃ + + * ┗━┓ ┏━┛ * ┃ ┃ + + + +Code is far away from * ┃ ┃ + bug with the animal protecting * ┃ ┗━━━┓ 神兽保佑,代码无bug * ┃ ┣┓ * ┃ ┏┛ * ┗┓┓┏━┳┓┏┛ + + + + * ┃┫┫ ┃┫┫ * ┗┻┛ ┗┻┛+ + + + */
echarts_部分图表配置_地图的配置
1 function echarts_map (thisId,data){ 2 $.get(\'map/json/henan.json\', function (chinaJson) { 3 echarts.registerMap(\'henan\', chinaJson); 4 var chart = echarts.init(document.getElementById(thisId)); 5 chart.setOption({ 6 7 visualMap: { 8 show:false,//不显示也有颜色区分的效果 9 min: 0, 10 max: 50000, 11 left: \'left\', 12 top: \'bottom\', 13 text: [\'高\',\'低\'], // 文本，默认为数值文本 14 calculable: false,//是否显示可拖动的句柄 15 inRange: { 16 color: [\'rgba(15,220,150,.5)\',\'#00415a\',] 17 } 18 }, 19 tooltip: { 20 trigger: \'item\', 21 showDelay: 0, 22 backgroundColor: \'#c6e8e9\',//浮框背景色 23 transitionDuration: 0.2, 24 formatter: function (params) {//设置提示框内的文字 25 var p = \'<h3 style="color:#29667b;text-align:center;font-size:14px;font-weight:bold;line-height:20px;">[\'+params.name+\']</h3>\'+ 26 \'<p style="color:#26677b;font-size:12px;">今日累计用户数：\'+params.data.uv+\'</p>\'+ 27 \'<p style="color:#26677b;font-size:12px;">当前活跃用户数：\'+params.data.iuv+\'</p>\' 28 return p; 29 } 30 }, 31 series: [{ 32 type: \'map\', 33 map: \'henan\', 34 roam: false,//标识是否可以缩放和拖动 35 aspectScale:1.1,//设置地图的长宽比 数值越大 高度越小 数值越小 宽度越小 36 zoom:1.25,//定义当前视角的缩放比 37 label: { 38 normal: { 39 show: false,//标示是否显示提示文字 40 }, 41 emphasis: { 42 show: false 43 } 44 }, 45 itemStyle:{ 46 normal:{ 47 areaColor:\'rgba(15,220,150,.5)\' 48 } 49 }, 50 data:data 51 }] 52 }); 53 chart.on(\'mouseover\', function (e) { 54 e.event.target.style.fill = \'#00415a\'; 55 }); 56 }); 57 };
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