蓝桥杯题库 历届试题部分（C++、Java）代码实现（1-15）
文章目录
- 五、历届试题
- PREV-1 核桃的数量
- PREV-2 打印十字图
- PREV-3 带分数
- PREV-4 剪格子
- PREV-5 错误票据
- PREV-6 翻硬币
- PREV-7 连号区间数
- PREV-8 买不到的数目
- PREV-9 大臣的旅费
- PREV-10 幸运数
- PREV-11 横向打印二叉树
- PREV-12 危险系数
- PREV-13 网络寻路
- PREV-14 高僧斗法
- PREV-15 格子刷油漆
将定期更新蓝桥杯习题集的解题报告~
五、历届试题
PREV-1 核桃的数量
#include import java.util.*;
public class Main {
public static void main(String[] args) {//
// TODO Auto-generated method stub
Scanner cin = new Scanner(System.in);
int[] arr = new int[3];
for (int i = 0;i<3;i++) {
arr[i] = cin.nextInt();
}
System.out.print(minPart(arr[0],minPart(arr[1],arr[2])));
cin.close();
}
private static int minPart(int i, int j) {//
int mul = i*j;
while(j!=0) {
int temp = j;
j = i % j;
i = temp;
}
return mul/i;
}
}
PREV-2 打印十字图
#include import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int N = scanner.nextInt();
int Len = (N+1)*2+1;
// 计算出右下角的图形
char[][] matrix = new char[Len][Len];
matrix[0][0] = matrix[0][1] = matrix[0][2] = '$';
// 计算第一行的值
for (int i = 3;i<Len;i+=2){
matrix[0][i] = '.';
matrix[0][i+1] = '$';
}
// 计算倒数第二行的值
for (int i = 0;i<Len;i++)
matrix[Len-2][i] = '.';
matrix[Len-2][Len-3] = '$';
// 计算倒数第一行的值
for (int i = 0;i<Len;i++)
matrix[Len-1][i] = '$';
matrix[Len-1][Len-1] = '.';
matrix[Len-1][Len-2] = '.';
// 计算中间行的值
for (int i = 1;i<Len-2;i++){
// 填充第 i 行的值
if (i%2==1){
// 如果是奇数行
int j = 0;
for (j = 0;j<i;j++){
matrix[i][j] = '.';
}
matrix[i][i-1] = '$';
matrix[i][j++] = '.';
matrix[i][j++] = '.';
matrix[i][j++] = '.';
for (int k = 0;k+j<Len;k++){
if (k%2==0)matrix[i][j+k] = '$';
else matrix[i][j+k] = '.';
}
} else {
// 如果是偶数行
int j = 0;
for (j = 0;j<i;j++){
matrix[i][j] = '$';
}
matrix[i][i-1] = '.';
matrix[i][j++] = '$';
matrix[i][j++] = '$';
matrix[i][j++] = '$';
for (int k = 0;k+j<Len;k++){
if (k%2==0)matrix[i][j+k] = '.';
else matrix[i][j+k] = '$';
}
}
}
for (int i = 0;i<Len;i++){
StringBuffer stringBuffer = new StringBuffer();
int hang = Len-1-i;
for (int j = 0;j<Len-1;j++){
stringBuffer.append(matrix[hang][Len-1-j]);
}
System.out.print(stringBuffer);
System.out.print(matrix[hang][0]);
System.out.println(stringBuffer.reverse());
}
for (int i = 1;i<Len;i++){
StringBuffer stringBuffer = new StringBuffer();
int hang = i;
for (int j = 0;j<Len-1;j++){
stringBuffer.append(matrix[hang][Len-1-j]);
}
System.out.print(stringBuffer);
System.out.print(matrix[hang][0]);
System.out.println(stringBuffer.reverse());
}
scanner.close();
}
}
PREV-3 带分数
简记：若强行跑O(11!)的时间复杂度再O(11)判断是否可行，会超时，需要在dfs过程中，通过维护已经得到的值来剪枝。
#include
import java.io.*;
import java.util.*;
class input {
private
static BufferedReader br =
new BufferedReader(new InputStreamReader(System.in));
private
static StringTokenizer tokenizer;
public
static String next() throws IOException {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
String line = br.readLine();
if (line == null) {
throw new IOException();
}
tokenizer = new StringTokenizer(line);
}
return tokenizer.nextToken();
}
public
static int nextInt() throws IOException { return Integer.parseInt(next()); }
} public class Main {
public
static void main(String[] args) throws IOException {
int n = input.nextInt();
int count = 0;
int sum = 0;
int[] dict = new int[100];
for (int i = 1; i < n; i++) {
for (int j = 1; j < 10000; j++) {
for (int k = 0; k < 10; k++) {
dict[k] = 0;
}
count = f(i, j, j * (n - i), dict);
if (count > 9) {
break;
}
if (dict[0] > 0 || count < 9) {
continue;
}
for (int k = 1; k < 10; k++) {
if (dict[k] != 1) {
count = -1;
break;
}
}
if (count == 9) {
sum++;
}
}
}
System.out.println(sum);
}
static int f(int a, int b, int c, int[] dict) {
int count = 0;
while (a > 0) {
dict[a % 10]++;
a /= 10;
count++;
}
while (b > 0) {
dict[b % 10]++;
b /= 10;
count++;
}
while (c > 0) {
dict[c % 10]++;
c /= 10;
count++;
}
return count;
}
}
PREV-4 剪格子
简记：一开始理解错了题意，后来百度的其他人方法，发现是一个玄学的dfs+剪枝，最差时间复杂度为：O(2^n*m)。认为没必要费时间练这题~
#include import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Main {
private static int M = 0, N = 0;
private static int sum = 0;
private static int[][] map = new int[10][10];
private static int[][] color = new int[10][10];
private static int min_unit_cnt = 0;
public static void main(String[] args) throws IOException {
BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
String str = bf.readLine();
String[] argNums = str.split(" ");
M = Integer.parseInt(argNums[0]);
N = Integer.parseInt(argNums[1]);
for (int i = 0;i < N;++i) {
String mapLine = bf.readLine();
String[] nums = mapLine.split(" ");
for (int j = 0;j < M;++j) {
map[i][j] = Integer.parseInt(nums[j]);
sum += map[i][j];
}
}
if (sum % 2 != 0) {
System.out.println(min_unit_cnt);
return;
}
dfs(0,0,0,0);
System.out.println(min_unit_cnt);
}
private static void dfs(int x, int y, int cnt, int dfs_sum) {
if (x >= N || y >= M || x < 0 || y < 0
|| (min_unit_cnt != 0 && cnt + 1 >= min_unit_cnt)
|| color[x][y] == 1
) {
return;
}
dfs_sum = dfs_sum + map[x][y];
if (dfs_sum > sum ) {
return ;
} else if (dfs_sum == sum / 2 ) {
if (min_unit_cnt == 0 || cnt + 1 < min_unit_cnt) {
min_unit_cnt = cnt + 1;
}
return;
} else {
color[x][y] = 1;
dfs(x, y+1,cnt+1,dfs_sum);
dfs(x+1, y,cnt+1,dfs_sum);
dfs(x, y-1,cnt+1,dfs_sum);
dfs(x-1, y,cnt+1,dfs_sum);
color[x][y] = 0;
return;
}
}
}
PREV-5 错误票据
简记：读入有些技巧，如果明白文件输入输出原理的话，会方便不少。
#include import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Scanner;
public class Main
{
public static void main(String[] args)
{
Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
int num = in.nextInt();
List<Integer> list = new ArrayList<Integer>();
for(int i = 0; i <= num; i++)
{
String[] strLine = in.nextLine().split(" ");
for(int j = 0; j < strLine.length; j++)
{
if(strLine[j].equals("")) continue;
list.add(Integer.parseInt(strLine[j]));
//System.out.println(list.get(list.size() - 1));
}
}
int min = 100010;
int max = -1;
int[] numLine = new int[100010];
for(int i = 0; i < list.size(); i++)
{
numLine[list.get(i)]++;
min = list.get(i) > min ? min : list.get(i);
max = list.get(i) > max ? list.get(i) : max;
}
int lost = 0;
int same = 0;
for(int i = min; i <= max; i++)
if(numLine[i] == 0) { lost = i; break;}
for(int i = min; i <= max; i++)
if(numLine[i] == 2) { same = i; break;}
out.println(lost + " " + same);
out.flush();
}
}
PREV-6 翻硬币
#include import java.io.*;
class Main{
static void change(int idx, StringBuilder cur){
String ex=cur.substring(idx, idx+2);
if(ex.equals("**")){
cur.replace(idx, idx+2, "oo");
}else if(ex.equals("*o")){
cur.replace(idx, idx+2, "o*");
}else if(ex.equals("o*")){
cur.replace(idx, idx+2, "*o");
}else{
cur.replace(idx, idx+2, "**");
}
}
public static void main(String[]args) throws IOException{
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
StringBuilder st = new StringBuilder(in.readLine());
String ed = in.readLine();
int n = st.length();
int step=0;
for(int i=0; i<n-1; i++){
char a = st.toString().charAt(i);
char b = ed.charAt(i);
if(a==b) continue;
else{
step+=1;
change(i, st);
}
}
System.out.println(step);
}
}
PREV-7 连号区间数
简记：n方暴力就可以过，没想到巧妙地枚举方法可以做到nlogn。
#include import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Main {
static int N = 100010;
static int[] f = new int[N];
public static void main(String[] args) throws NumberFormatException, IOException {
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(reader.readLine().trim());
String[] s1 = reader.readLine().split(" ");
for(int i = 1;i <= n;i++) f[i] = Integer.parseInt(s1[i - 1]);
int res = 0;
for(int i = 1;i <= n;i++)
{
int minv = Integer.MAX_VALUE;
int maxv = Integer.MIN_VALUE;
for(int j = i;j <= n;j++)
{
minv = Math.min(minv, f[j]);
maxv = Math.max(maxv, f[j]);
if(maxv - minv == j - i) res ++;
}
}
System.out.println(res);
}
}
PREV-8 买不到的数目
题目大意：输入两个数：a,b，找出最大的c满足c不能表示成an+bm(n,m∈N*)
思路：打表找规律
#include
{
int c = a;
a = b;
b = c;
}
int maxx = 0;
for (int i = 1; i < b; i++) {
int x = i;
while (1) {
if (x % a == 0) break;
x += b;
}
// cout<
if (maxx < x) maxx = x;
}
cout << maxx - b;
return 0;
}
import java.io.BufferedReader;
import java.io.InputStreamReader;
public class Main {
public static void main(String[] args) throws Throwable {
BufferedReader buf = new BufferedReader(new InputStreamReader(System.in));
String strNum = buf.readLine();
String[] num = strNum.split(" ");
int a, b;
a = Integer.valueOf(num[0]);
b = Integer.valueOf(num[1]);
if (a > b) {
int tem = a;
a = b;
b = tem;
}
int c = a * b;
int tem = c;
while (tem > 0) {
if (tem % a == 0)
tem = --c;
else if (tem % b == 0)
tem = --c;
else
tem -= b;
}
System.out.println(c);
}
}
PREV-9 大臣的旅费
简记：
1.问题等价于求一棵树的重心
2.实测点数n<=10000,最大路费不超过signed int。
#include import java.io.BufferedInputStream;
import java.io.IOException;
import java.util.ArrayList;
public class Main
{
private static BufferedInputStream in = new BufferedInputStream(System.in);
private static ArrayList<Node> n = new ArrayList<Node>();
private static Integer dis = 0;
private static Integer pow = 0;
public static void main(String[] args) throws IOException
{
int size = readInt();
for(int i=0; i<size; i++)
{
n.add(null);
}
for(int i=1; i<size; i++)
{
int x = readInt()-1;
int y = readInt()-1;
int d = readInt();
Node node = new Node();
node.nextSide = n.get(x);
node.con = y;
node.power = d;
n.set(x, node);
node = new Node();
node.nextSide = n.get(y);
node.con = x;
node.power = d;
n.set(y, node);
}
away(0, 0, -1);
pow = 0;
away(dis, 0, -1);
System.out.println(pow * 10 + (1 + pow) * pow /2);
}
private static void away(int index, int power, int from)
{
if(pow < power)
{
pow = power;
dis = index;
}
Node node = n.get(index);
while(node != null)
{
if(node.con == from)
{
node = node.nextSide;
continue;
}
away(node.con, power + node.power, index);
node = node.nextSide;
}
}
private static int readInt() throws IOException
{
int i,sum=0;
while(((i=in.read())&48) != 48 || i>57);
for(;(i&56) == 48 || (i&62) == 56; i=in.read())
sum = sum*10 + (i&15);
return sum;
}
private static class Node
{
int power;
int con;
Node nextSide;
}
}
PREV-10 幸运数
简记：
1.一般这种求区间[L,R]之间个数的，变成求[1,R]个数-[1,L-1]个数。
2.乍一看，平均下来每次压缩是个logn级别的操作,权当一试。
3.1e6超时，5e5优化到了极致，可以通过数据，但是应该还是有问题，我觉得是题目问题。
#include import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int m = sc.nextInt();
int n = sc.nextInt();
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = 2 * i + 1;
}
int l = 1;// 幸运数下标
while (true) {
int p = l + 1;// 指向数字向前移动的位置
for (int i = l + 1; i < n; i++) {
if ((i + 1) % a[l] != 0) {
a[p] = a[i];
p++;
}
if (a[p] > n)
break;
}
l++;
if (a[l] >= n)
break;
}
int count = 0;
for (int i = 0; i < n; i++) {
if (a[i] >= n)
break;
if (a[i] > m)
count++;
}
System.out.println(count);
}
}
PREV-11 横向打印二叉树
简记：
这次练一下指针形式建树。这个横向输出实在麻烦，预声明了二维字符数组，递归画出，最后输出。
#include import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class Main {
static int n;
static int[] a=new int[108];
static int[] left;
static int[] right;
static int[] level;
static int maxl=0;
static int M,N;
static char[][] map;
static boolean[] flag;
static boolean[] flag2;
static void add(int root,int index)
{
if(a[index]<a[root])
if(left[root]==0) { left[root]=index; level[index]=level[root]+1; if(level[index]>maxl) maxl=level[index]; }
else add(left[root],index);
else
if(right[root]==0) { right[root]=index; level[index]=level[root]+1; if(level[index]>maxl) maxl=level[index]; }
else add(right[root],index);
}
static void Fill(int root,int x,int y)
{
int lvl=level[root]+1;
String num=String.valueOf(a[root]);
int len=num.length();
flag[x]=true;
for(int i=0;i<len;i++)
{
map[x][y-i]=num.charAt(i);
int h=2<<(maxl-lvl);
for(int j=1;j<h;j++) { map[x+j][y-i]='.'; map[x-j][y-i]='.'; }
}
if(left[root]==0 && right[root]==0 && level[root]<maxl) flag2[x]=true;
if(left[root]!=0)
Fill(left[root],x+(2<<(maxl-lvl))/2,y-8);
if(right[root]!=0)
Fill(right[root],x-(2<<(maxl-lvl))/2,y-8);
}
public static void main(String[] args) throws IOException {
BufferedReader bfr = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer tok=new StringTokenizer(bfr.readLine());
// long begin=System.currentTimeMillis();
int i,j;
for(i=1;tok.hasMoreTokens();i++) a[i]=Integer.parseInt(tok.nextToken());
n=i;
left =new int[n];
right =new int[n];
level=new int[n];
left[0]=1; right[0]=1; level[0]=0; level[1]=1;
for(i=2;i<n;i++)
add(1,i);
M=(2<<(maxl-1))-1;
N=8*maxl-3;
map=new char[M][N];
flag=new boolean[M];
flag2=new boolean[M];
for(int l=1;l<maxl;l++)
{
int ty=l*8-1;
int start=(2<<(l-1))-1,d=(2<<l);
for(i=0;i<M;i++) map[i][ty]=map[i][ty-1]=map[i][ty-2]='.';
for(i=start;i<M;i+=d)
{
map[i][ty]='-';
int hei=(2<<(l-1))/2;
for(int ii=0;ii<=hei;ii++)
{
map[i+ii][ty-1]='|';
map[i-ii][ty-1]='|';
}
map[i+hei][ty-2]='-';
map[i-hei][ty-2]='-';
}
}
Fill(1,M/2,N-1);
for(i=0;i<M;i++)
{
if(flag[i])
{
for(j=0;;j++) if(map[i][j]!=0 && map[i][j]!='.') break;
int jj;
for(jj=N-1;jj>j+1;jj--) if(map[i][jj]!=0) System.out.print(map[i][jj]);
if(!flag2[i]) System.out.print(map[i][jj]+""+map[i][jj-1]);
System.out.println();
}
}
// System.out.println(System.currentTimeMillis()-begin);
}
}
PREV-12 危险系数
简记：
依次在图中删除每一个点，判断s到t是否依然连通，时间复杂度O(n*m)。直觉感觉应该是还有更快的办法，没认真想。
#include import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.StringTokenizer;
public class Main {
static int n,m;
public static void main(String[] args) throws IOException {
BufferedReader bfr=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer tok=new StringTokenizer(bfr.readLine());
// long begin=System.currentTimeMillis();
n=Integer.parseInt(tok.nextToken());
m=Integer.parseInt(tok.nextToken());
ArrayList<Integer>[] way=new ArrayList[n];
int count=0;
int i;
for(i=0;i<n;i++) way[i]=new ArrayList<Integer>();
for(i=0;i<m;i++)
{
tok=new StringTokenizer(bfr.readLine());
int x=Integer.parseInt(tok.nextToken())-1, y=Integer.parseInt(tok.nextToken())-1;
way[x].add(y); way[y].add(x);
}
tok=new StringTokenizer(bfr.readLine());
int g1=Integer.parseInt(tok.nextToken())-1, g2=Integer.parseInt(tok.nextToken())-1;
for(i=0;i<n;i++)
{
if(i==g1 || i==g2) continue;
boolean[] flag=new boolean[n];
int[] que=new int [n];
int front=0,rear=1;
que[0]=g1;
while(front<rear)
{
for(int p=0;p<way[que[front]].size();p++)
{
if(way[que[front]].get(p)==i) continue;
if (!flag[way[que[front]].get(p)])
{
que[rear] = way[que[front]].get(p);
flag[que[rear]]=true;
if(que[rear]==g2) break;
rear++;
}
}
if(flag[g2]) { count++; break; }
front++;
}
}
System.out.println(n-2-count);
// System.out.println(System.currentTimeMillis()-begin);
}
}
PREV-13 网络寻路
简记：
枚举每条边(u,v)，以这条边为中间边的合法边数为：(和u相连的点数-1)*(和v相连的点数-1)
#include
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.StringTokenizer;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader bfr = new BufferedReader(
new InputStreamReader(System.in));
StringTokenizer tok = new StringTokenizer(bfr.readLine());
int n = Integer.parseInt(tok.nextToken());
int m = Integer.parseInt(tok.nextToken());
int i, count = 0;
ArrayList<Integer>[] way = new ArrayList[n];
for (i = 0; i < n; i++)
way[i] = new ArrayList<Integer>();
for (i = 0; i < m; i++) {
tok = new StringTokenizer(bfr.readLine());
int x = Integer.parseInt(tok.nextToken()) - 1;
int y = Integer.parseInt(tok.nextToken()) - 1;
way[x].add(y);
way[y].add(x);
}
for (i = 0; i < n; i++) {
int a = i, ll = way[i].size();
for (int ii = 0; ii < ll; ii++) {
int aa = way[a].get(ii);
int lll = way[aa].size();
for (int iii = 0; iii < lll; iii++) {
int aaa = way[aa].get(iii);
if (aaa == a)
continue;
count += way[aaa].size();
count--;
}
}
}
System.out.println(count);
}
}
PREV-14 高僧斗法
简记：
首先两两组合看成nim博奕问题，奇数个就在最高点加一个。但是目标是一定要优先移动最左边的。有个坑点，对于一对点，可以变大也可以变小。
#include import java.util.Scanner;
public class Main{
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc=new Scanner(System.in);
String A=sc.nextLine();
String[] arrayA=A.split(" ");
int[] a=new int[arrayA.length];
for(int i=0;i<a.length;i++){
a[i]=Integer.valueOf(arrayA[i]);
}
//传入的数组a为小和尚所站的位置
solve(a);
}
//任何无偏的游戏理论上可以转换为尼姆堆
private static void solve(int[] a) {
// TODO Auto-generated method stub
for(int i=0;i<a.length-1;i++)
for(int k=a[i]+1;k<a[i+1];k++){
int old=a[i];
try{
a[i]=k;//试探
if(f(a)==false){
System.out.println(old+" "+k);
return;
}
}finally{
a[i]=old;//回溯
}
}
}
private static boolean f(int[] a) {
// TODO Auto-generated method stub
int sum=0;
for(int i=0;i<a.length-1;i+=2){
sum^=(a[i+1]-a[i]-1);
}
return sum!=0;
}
}
PREV-15 格子刷油漆
简记：动态规划
#include import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scan=new Scanner(System.in);
int n=scan.nextInt();
long a[]=new long[n+1];
long b[]=new long[n+1];
long sum;
b[1]=1;
for(int i=2;i<=n;i++){
b[i]=2*b[i-1]%1000000007;
}
a[1]=1;a[2]=6;
for(int i=3;i<=n;i++){
a[i]=(2*a[i-1]+b[i]+4*a[i-2])%1000000007;
}
sum=(4*a[n])%1000000007;
for(int i=2;i<n;i++){
sum+=((8*b[i-1]*a[n-i])%1000000007+(8*b[n-i]*a[i-1])%1000000007)%1000000007;//必须每个项都取余，防止有大于这个数的情况
sum%=1000000007;
}
System.out.println(sum);
}
}