第十五届蓝桥杯大赛软件赛省赛javaB组（蒟蒻赛时写的题

4个小时，下来感觉，迷迷糊糊的（感觉写的题和没写一样，要么暴力，要么写写if els,感觉没有什么体验感。前两个填空题都是签到，第二个填空写的暴力，感觉跑了5分钟左右，其实是可以前缀和优化的，第三个题就不会那个输入了，后面迷糊的有点忘了，用的学校机房的电脑（悲，用着不太习惯，到9点整才解除电脑控屏，到那时我才把那个idea,加载数据qwq..还要调式那个idea,

第一题：求20 24的倍数的第202420242024项是什么？

数学：其实就是除2乘24，

private static void solve()throws IOException {long n=24,m=101210121012L;pw.println(n*m);}

第二题：

题意不好读，其实题意大概是假设一个数字n其数位有k,用前k位累加就是当前的一位的数值，一开始的k位，是n这个数字按位拆开，赛时写的代码依托答辩，

import java.util.*;
import java.io.*;public class Main {static int N=(int)1e6+7;static int[] a;static int n,m,q,k;static List<Integer>[] g=new ArrayList[N];static boolean f(int n){int cnt=0,t=n;while(t>0){cnt++;t/=10;}int[] nums=new int[N];t=0;int m=n;while(m>0){t=t*10+m%10;m/=10;}for(int i=0;i<cnt;i++){nums[i]=t%10;t/=10;}for(int i=cnt;;i++){for(int j=i-1,c=cnt;c>0;c--,j--){nums[i]+=nums[j];}if(nums[i]==n)  {for(int j=0;j<=i;j++){pw.print(nums[j]+" ");}pw.println();return true;}if(nums[i]>n){return false;}}}private static void solve()throws IOException {int mx=(int)1e7;for(int i=mx;i>=0;i--){if(f(i)){pw.println(i);break;}}/*7 9 1 3 8 3 7 38 69 129 257 511 1014 2025 4043 8048 16027 31925 63593 126675 252336 502647 1001251 1994454 3972881 7913837
7913837*/}public static void main(String[] args)throws IOException {int T=1;for(int i=1;i<=T;i++){solve();}pw.flush();pw.close();}static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));static StreamTokenizer st=new StreamTokenizer(br);static PrintWriter pw=new PrintWriter(new OutputStreamWriter(System.out));static int nextInt()throws IOException{st.nextToken();return (int)st.nval;}static int nextLong()throws IOException{st.nextToken();return (int)st.nval;}}

第三题：

按题意模拟即可，不会输入的哭死，不太清楚下面的代码，输入有没有问题

//package A;import java.util.*;
import java.io.*;public class Main {static int N=(int)2e5+7,INF=0x3f3f3f3f;static int[] a;static int n,m,q,k;static List<Integer>[] g=new ArrayList[N];private static void solve()throws IOException {n=sc.nextInt();
//        pw.println(n);  pw.flush();a=new int[n];//        for(int i=0;i<11;i++)while(sc.hasNext()){String s=sc.next();if(s.equals("add")) {int k=sc.nextInt();a[0]++;}else if(s.equals("query")){int ans=a[0];for(int j=0;j<n;j++)   ans=Math.min(ans,a[j]);System.out.println(ans);}else if(s.equals("sync")) {int k=sc.nextInt();if(a[k]<a[0])   a[k]++;}}}/*
3
add 1
add 2
query
add 1
sync 1
sync 1
sync 2
query
sync 1
query
sync 2
sync 2
sync 1
query*/public static void main(String[] args)throws IOException {int T=1;for(int i=1;i<=T;i++){solve();}
//        pw.flush();
//        pw.close();}static Scanner sc=new Scanner(System.in);static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));static StreamTokenizer st=new StreamTokenizer(br);static PrintWriter pw=new PrintWriter(new OutputStreamWriter(System.out));static int nextInt()throws IOException{st.nextToken();return (int)st.nval;}static int nextLong()throws IOException{st.nextToken();return (int)st.nval;}}

第四题：

发现暴力不太会写，写的单次查询o1,但是写到后面感觉有太多情况了，这if else太吓人了，，，

直接草草收场，交了

//package A;import java.util.*;
import java.io.*;public class Main {static int N=(int)2e5+7,INF=0x3f3f3f3f;static int[] a;static int n,m,q,k;static List<Integer>[] g=new ArrayList[N];private static void solve()throws IOException {int n=nextInt();for(int i=0;i<n;i++){int c2=nextInt(),c3=nextInt(),c4=nextInt();int b4=nextInt(),b6=nextInt();int ans=0,k3=c3/2;  //三人凑6人桌if(b6>=k3){b6-=k3;c3-=k3*2;ans+=k3*6;}else{ans+=b6*6;c3-=2*b6;b6=0;}int mi=Math.min(c2,c4);if(b6>=mi){ //在考虑2 4人凑6人桌b6-=mi;c2-=mi;c4-=mi;ans+=6*mi;}else{c2-=b6;c4-=b6;ans+=b6*6;b6=0;}if(b4>=c4){ //4人,安排4人桌子ans+=c4*4;b4-=c4;c4=0;}else{ans+=b4*4;c4-=b4;b4=0;}int k2=c2/2;if(b4>=k2){ //二分凑 4卓b4-=k2;ans+=k2*4;c2-=2*k2;}else{ans+=b4*4;c2-=2*b4;b4=0;}if(b4>=c3){ans+=3*c3;b4-=c3;c3=0;}else{ans+=b4*3;c3-=b4;b4=0;}int k=b4*2+b6*3;ans+=Math.min(k,c2);pw.println(ans);}/*23 0 1 0 10 2 2 1 1610*/}public static void main(String[] args)throws IOException {int T=1;for(int i=1;i<=T;i++){solve();}pw.flush();pw.close();}static Scanner sc=new Scanner(System.in);static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));static StreamTokenizer st=new StreamTokenizer(br);static PrintWriter pw=new PrintWriter(new OutputStreamWriter(System.out));static int nextInt()throws IOException{st.nextToken();return (int)st.nval;}static int nextLong()throws IOException{st.nextToken();return (int)st.nval;}}

第五题：

直接看不懂中文了，瞎写的

第六题：

点的数量是1000，边的数量最多是5000，询问次数是5e4,对每一个询问，bfs求边数小于lim能到达的所有结点，时间复杂度是q(n+m)=5e4*6e3=3e8,好像差不多可以过，不可以过那么直接就是寄了

，最后用每一次询问可以到达的结点累加起来，再除一个n就是答案了

//package A;import java.util.*;
import java.io.*;public class Main {static int N=(int)1e3+7;static int[] a,d=new int[N],q=new int[N];static int n,m,q1,k;static List<Integer>[] g=new ArrayList[N];static boolean[] vis=new boolean[N];static int bfs(int start,int limit){for(int i=1;i<=n;i++)   d[i]=-1;int tt=-1,hh=0;q[++tt]=start;d[start]=0;while(tt>=hh){int x=q[hh++];//            pw.print(x+" ");for(int y:g[x]){if(d[y]==-1&&d[x]+1<=limit){q[++tt]=y;d[y]=d[x]+1;}}}
//        pw.println();return tt+1;}private static void solve()throws IOException {for(int i=0;i<N;i++)    g[i]=new ArrayList<>();n=nextInt(); m=nextInt(); q1=nextInt();for(int i=0;i<m;i++){int x=nextInt(),y=nextInt();g[x].add(y);g[y].add(x);}double ans=0;for(int i=0;i<q1;i++){int st=nextInt(),lim=nextInt();ans+=bfs(st,lim);}pw.printf("%.2f",ans/n);}/*3 2 31 22 32 12 01 1*/public static void main(String[] args)throws IOException {int T=1;for(int i=1;i<=T;i++){solve();}pw.flush();pw.close();}static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));static StreamTokenizer st=new StreamTokenizer(br);static PrintWriter pw=new PrintWriter(new OutputStreamWriter(System.out));static int nextInt()throws IOException{st.nextToken();return (int)st.nval;}static int nextLong()throws IOException{st.nextToken();return (int)st.nval;}}

第七题：

写的暴力+剪枝，对每一个字母枚举是否可以旋转，可能有两种或者4种情况，然后再对每一次旋转情况，做一边dfs,判断是否可以包含4个字母，感觉写dfs的话就是码农题了，不知道这个题可以不可以状态压缩dp,QAQ

//package A;import java.util.*;
import java.io.*;public class Main {static int N=57;static int[][] a=new int[N][N],b=new int[N][N];static int n,m,q1,k;static boolean[] vis=new boolean[N];static int[][] d1x=new int[4][];static int[][] d2x=new int[4][];static int[][] d3x=new int[4][];static int[][] d4x=new int[4][];static int[][] d1y=new int[4][];static int[][] d2y=new int[4][];static int[][] d3y=new int[4][];static int[][] d4y=new int[4][];static int[][] dx=new int[4][],dy=new int[4][];static boolean ok;static void dfs(int x,int y,int dep){if(dep==4||ok){ok=true;return;}if(y==n){x++;y=0;}if(x==n)    return;dfs(x,y+1,dep);if(a[x][y]==1){boolean flag=true;int k=4;if(dep==1||dep==3)   k=2;for(int i=0;i<k;i++){int bx=dx[dep][i]+x,by=dy[dep][i]+y;if(bx<0||by<0||bx>=n||by>=n||a[bx][by]!=1){flag=false;break;}}if(!flag)   return;for(int i=0;i<k;i++){   //改int bx=dx[dep][i]+x,by=dy[dep][i]+y;a[bx][by]=2;}dfs(x,y+1,dep+1);for(int i=0;i<k;i++){   //复原int bx=dx[dep][i]+x,by=dy[dep][i]+y;a[bx][by]=b[bx][by];}}}private static void solve()throws IOException {n=nextInt();for(int i=0;i<n;i++)for(int j=0;j<n;j++)   {a[i][j]=nextInt();b[i][j]=a[i][j];}for(int i1=0;i1<3;i1++)for(int i2=0;i2<2;i2++)for(int i3=0;i3<2;i3++)for(int i4=0;i4<3;i4++){dx[0]=d1x[i1];dx[1]=d2x[i2];dx[2]=d3x[i3];dx[3]=d4x[i4];dy[0]=d1y[i1];dy[1]=d2y[i2];dy[2]=d3y[i3];dy[3]=d4y[i4];for(int i=0;i<n;i++)for(int j=0;j<n;j++)a[i][j]=b[i][j];ok=false;dfs(0,0,0);if(ok){pw.println("Yes");return;}}pw.println("No");}/*
2
5
1 1 1 1 1
1 0 1 1 0
1 0 0 0 1
1 0 1 0 1
1 1 1 1 1
5
1 0 0 1 1
1 1 1 1 1
1 1 1 1 0
1 1 1 0 1
0 1 1 1 1*/public static void main(String[] args)throws IOException {//1d1x[0]=new int[]{0,1,2,2};d1y[0]=new int[]{0,0,0,1};d1x[1]=new int[]{0,0,0,1};d1y[1]=new int[]{0,1,2,0};d1x[2]=new int[]{0,0,1,2};d1y[2]=new int[]{0,1,1,1};d1x[3]=new int[]{0,1,1,1};d1y[3]=new int[]{0,0,-1,-2};//2d2x[0]=new int[]{0,1,2,3};d2y[0]=new int[]{0,0,0,0};d2x[1]=new int[]{0,0,0,0};d2y[1]=new int[]{0,1,2,3};//3d3x[0]=new int[]{0,0,0,1};d3y[0]=new int[]{0,1,2,1};d3x[1]=new int[]{0,1,2,1};d3y[1]=new int[]{0,0,0,-1};d3x[2]=new int[]{0,1,1,1};d3y[2]=new int[]{0,0,-1,1};d3x[3]=new int[]{0,1,2,1};d3y[3]=new int[]{0,0,0,1};//4d4x[0]=new int[]{0,0,1,1};d4y[0]=new int[]{0,1,0,-1};d4x[1]=new int[]{0,1,1,2};d4y[1]=new int[]{0,0,1,2};int T=nextInt();for(int i=1;i<=T;i++){solve();}pw.flush();pw.close();}static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));static StreamTokenizer st=new StreamTokenizer(br);static PrintWriter pw=new PrintWriter(new OutputStreamWriter(System.out));static int nextInt()throws IOException{st.nextToken();return (int)st.nval;}static int nextLong()throws IOException{st.nextToken();return (int)st.nval;}}

第八题：

写的O(n^2)暴力o.O,,其实可以分颜色暴力的，但是写完上面那个码农提，脑子拓机了

package A;import java.util.*;
import java.io.*;public class Main {static int N=(int)1e5+7,mod=(int)1e9;static int[] a=new int[N],b=new int[N],c=new int[N];static int n,m,q1,k;static List<Integer>[] g=new ArrayList[N];static void dfs(int x,int y){}private static void solve()throws IOException {n=nextInt();for(int i=0;i<n;i++){a[i]=nextInt();b[i]=nextInt();c[i]=nextInt();}long ans=0;for(int i=0;i<n;i++){for(int j=0;j<i;j++){if(c[i]==c[j])  continue;if(a[i]>b[j]&&b[i]<a[j]||(a[i]<b[j]&&b[i]>a[j]))    ans++;}}pw.println(ans%mod);}/*51 10 06 6 08 6 16 10 01 2 1*/public static void main(String[] args)throws IOException {int T=1;for(int i=1;i<=T;i++){solve();}pw.flush();pw.close();}static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));static StreamTokenizer st=new StreamTokenizer(br);static PrintWriter pw=new PrintWriter(new OutputStreamWriter(System.out));static int nextInt()throws IOException{st.nextToken();return (int)st.nval;}static int nextLong()throws IOException{st.nextToken();return (int)st.nval;}}

总结一下：java还是不会那个不确定个数的输入，赛时一直再调，不太习惯写java,当然主要原因还是因为菜（菜是原罪，思路跟不上手速，就写下这么多吧，等下要去看奥德曼打怪兽了（哥斯拉2，QWQ

预祝大家都是省一啊o.O,o.O,