蓝桥杯题库 历届试题部分（C++、Java）代码实现（31-45）
文章目录
- 五、历届试题
- PREV-31 小朋友排队
- PREV-32 分糖果
- PREV-33 兰顿蚂蚁
- PREV-34 矩阵翻硬币
- PREV-35 正则问题
- PREV-36 包子凑数
- PREV-37 分巧克力
- PREV-38 油漆面积
- PREV-39 日期问题
- PREV-40 k倍区间
- PREV-41 Excel地址
- PREV-42 九宫幻方
- PREV-43 拉马车
- PREV-44 青蛙跳杯子
- PREV-45 图形排版
将定期更新蓝桥杯习题集的解题报告~
五、历届试题
PREV-31 小朋友排队
简记：树状数组维护
设：
f(i)=a[i]左边严格大于a[i]的元素个数。
g(i)=a[i]右边严格小于a[i]的元素个数。
则：
a[i]的不高兴度为：1+2+…+f(i) + 1+2+…+g(i);
#include import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Main{
static int N = 100010;
static int MAX = 1000100;
static int[] C = new int[MAX];
static int[] S = new int[MAX];
static int[] b = new int[N];
static long[] total = new long[N];
static long ans;
static int[] num = new int[N];
static int T, s, t, i, j;
static int Lowbit(int x) {
return x & (-x);
}
static void add(int pos, int num, int[] P) {
while (pos <= MAX) {
P[pos] += num;
pos += Lowbit(pos);
}
}
static int Sum(int end, int[] P) {
int cnt = 0;
while (end > 0) {
cnt += P[end];
end -= Lowbit(end);
}
return cnt;
}
static void init() {
total[0] = 0;
for (int i = 1; i < N; ++i) {
total[i] = total[i - 1] + i;
}
}
public static void main(String[] args) throws IOException {
init();
BufferedReader buf = new BufferedReader(
new InputStreamReader(System.in));
T = Integer.parseInt(buf.readLine());
String[] str = buf.readLine().split(" ");
for (int j = 0; j < T; j++) {
num[j] = Integer.parseInt(str[j]);
add(num[j] + 1, 1, C);
b[j] = j - Sum(num[j], C);
b[j] -= Sum(num[j] + 1, C) - Sum(num[j], C) - 1;
}
ans = 0;
for (int j = T - 1; j > -1; --j) {
add(num[j] + 1, 1, S);
b[j] += Sum(num[j], S);
ans += total[b[j]];
}
System.out.println(ans);
}
}
PREV-32 分糖果
#include
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Main
{
static int m,n;
public static void main(String[] args) throws IOException
{
int candy=0;
BufferedReader buf;
String str1,str2;
buf=new BufferedReader(new InputStreamReader(System.in));
str1=buf.readLine();
m=Integer.parseInt(str1);
int b[]=new int[m];
str2=buf.readLine();
String a[]=str2.split(" ");
for(int i=0;i<m;i++)
{
b[i]=Integer.parseInt(a[i]);
}
if(!equal(b))
{
while(true)
{
n=b[m-1];
for(int i=m-1;i>0;i--)
{
b[i]=b[i-1]/2+b[i]/2;
}
b[0]=b[0]/2+n/2;
if(equal(b))
break;
else
{
for(int i=0;i<m;i++)
{
if(b[i]%2!=0)
{
candy++;
b[i]++;
}
}
}
}
}
System.out.println(candy);
}
public static boolean equal(int a[])
{
int num=0;
for(int i=0;i<m-1;i++)
{
if(a[i]==a[i+1])
num++;
}
if(num==m-1)
{
return true;
}
else
return false;
}
}
PREV-33 兰顿蚂蚁
//简单模拟
#include
for (int i = 0; i < k; i++) {
move();
}
cout << y << " " << x;
return 0;
}
import java.io.*;
public class Main {
static int n, m;
static int s, e;
static char[] chs = { 'L', 'U', 'R', 'D', 'L' };
static int count = 0;
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s1[] = br.readLine().split(" ");
n = Integer.parseInt(s1[0]);
m = Integer.parseInt(s1[1]);
int[][] arr = new int[n][m];
for (int a = 0; a < n; a++) {
String str[] = br.readLine().split(" ");
for (int b = 0; b < m; b++) {
arr[a][b] = Integer.parseInt(str[b]);
}
}
int x, y;
String s2[] = br.readLine().split(" ");
x = Integer.parseInt(s2[0]);
y = Integer.parseInt(s2[1]);
char dec = s2[2].charAt(0);
int z;
z = Integer.parseInt(s2[3]);
s = x;
e = y;
while (count < z) {
if (arr[s][e] == 1) {
for (int i = 0; i < chs.length; i++) {
if (dec == chs[i]) {
dec = chs[i + 1];
break;
}
}
arr[s][e] = 0;
func(dec, s, e);
}
if (arr[s][e] == 0) {
for (int j = 1; j < chs.length; j++) {
if (dec == chs[j]) {
dec = chs[j - 1];
break;
}
}
arr[s][e] = 1;
func(dec, s, e);
}
}
System.out.print(s + " " + e);
}
public static void func(char dec, int x, int y) {
if (dec == 'L') {
e -= 1;
count++;
}
if (dec == 'U') {
s -= 1;
count++;
}
if (dec == 'R') {
e += 1;
count++;
}
if (dec == 'D') {
s += 1;
count++;
}
}
}
PREV-34 矩阵翻硬币
简记：
思路：m*n的方格，ans = sqrt(m)*sqrt(n); 向下取整floor。大数据，二分求根号。
注：模拟十进制只能90分，压万进制可以满分
#include
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.StringTokenizer;
public class Main {
public static void main(String[] args) throws Exception {
/*n = Reader_09.nextInt();
m = Reader_09.nextInt();*/
//由于数目很大,所以不能用netxInt() 所以用String 最后转换成bigInteger
String n = Reader_10.next();
String m = Reader_10.next();
System.out.println(sqrt(n).multiply(sqrt(m)));
}
private static BigInteger sqrt(String s) {
//存一下数字的长度
int length = s.length();
//操作的位数
int len = 0;
//偶数
if (length % 2 == 0) {
len=length/2;
}else {
len=length/2+1;
}
char[] sArr = new char[len];
//将字符数组默认位数上的数都填充为0
Arrays.fill(sArr, '0');
//转换成大整数
BigInteger target = new BigInteger(s);
for (int pos = 0; pos < len; pos++) {
//从1~9开始尝试
for (char c = '1'; c <= '9'; c++) {
sArr[pos] = c;//在pos这个位置上尝试
//转换成字符串,填入biginteger ,并用bigInteger的性质平方
BigInteger pow = new BigInteger(String.valueOf(sArr)).pow(2);
//大整数的比较方法
if (pow.compareTo(target) == 1) {
//如果用2超出了,就填1,如果8超出了,就填7
sArr[pos] -= 1;
break;
}
}
}
return new BigInteger(String.valueOf(sArr));
}
}
class Reader_10 {
static BufferedReader br = new BufferedReader(new InputStreamReader(
System.in));
static StringTokenizer tokenizer = new StringTokenizer("");
static String next() throws Exception {
while (!tokenizer.hasMoreTokens()) {
tokenizer = new StringTokenizer(br.readLine());
}
return tokenizer.nextToken();
}
static int nextInt() throws Exception {
return Integer.parseInt(next());
}
}
PREV-35 正则问题
简记：
对于解决带括号的表达式求解问题，递归比较方便。
#include import java.util.Scanner;
public class Main {
static int position =0;
static char [] a;
public static void main(String[] args) {
Scanner in =new Scanner (System.in);
String str=in.nextLine();
a=str.toCharArray();
in.close();
System.out.println(read());
}
static int read() {
int max=0;
int temp=0;
while(position<a.length) {
if(a[position]=='(') {position++;max+=read();continue;}
if(a[position]=='x') {position++;max++;continue;}
if(a[position]=='|') {position++;temp=max(max,temp);max=0;continue;}
if(a[position]==')') {position++;return max(max,temp);}
}
return max(max,temp);
}
private static int max(int max, int temp) {
if(max>=temp)return max;
else return temp;
}
}
PREV-36 包子凑数
简记：数论常识+打表找规律
#include
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n =in.nextInt();//有几个笼子
int[] a = new int[n];//存储每个笼子能够装多少包子
int[] number=new int[10001];//存储能够凑齐的包子数量
int count=0;
for (int i=0;i<n;i++){
a[i]=in.nextInt();
number[a[i]]=1;
}
int g=a[0];
for (int i =1;i<a.length;i++){
g=gcd(g,a[i]);
}
if (g!=1){
System.out.println("INF");
}else {
for (int i=0;i<a.length;i++){
for (int j=0;j+a[i]<10001;j++){
if (number[j]==1) number[j+a[i]]=1;
}
}
for (int i=0;i<10001;i++){
if (number[i]==0) count++;
}
System.out.println(count-1);//减去number[0]的情况
}
}
//求最大公约数
public static int gcd(int a,int b){
return b==0?a:gcd(b,a%b);
}
}
PREV-37 分巧克力
简记：整体二分答案
#include import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class Main {
static class InputReader{
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream),32768);
tokenizer = null;
}
public String next() {
while(tokenizer == null || !tokenizer.hasMoreElements()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
}catch(IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
public static void main(String[] args) {
InputReader s = new InputReader(System.in);
int n = s.nextInt();
int k = s.nextInt();
int h[] = new int[n];
int w[] = new int[n];
for(int i=0;i<n;i++) {
h[i] = s.nextInt();
w[i] = s.nextInt();
}
int l=1,r=100000,ans=1;
while(l<=r) {
int mid = (l+r)/2;
int t = 0;
for(int i=0;i<n;i++) {
t+=(h[i]/mid)*(w[i]/mid);
if(t>=k) {
l = mid+1;
ans = mid;
continue;
}
}
if(t<k) {
r = mid-1;
}
}
System.out.println(ans);
}
}
PREV-38 油漆面积
简记：本来是线段树扫描线的，暴力冲过去了
#includePREV-39 日期问题
简记：模拟
#include import java.util.Scanner;
import java.util.TreeSet;
public class Main {
/**
* @param args
* @return
*/
public static java.lang.String rq(int a, int b, int c){
if(a>=0 && a<=59) a = a + 2000;
else if(a>=60 && a<=99) a = a + 1900;
else
return "";
if(b<1 || b>12)
return "";
if(c<1 || c>31)
return "";
switch(b){
case 2:
boolean temp = run(a);
if(temp){
if(c>29) return "";}
else
if(c>28)
return "";
break;
case 4:
if(c>30) return "";
break;
case 6:
if(c>30) return "";
break;
case 9:
if(c>30) return "";
break;
case 11:
if(c>30) return "";
break;
}
String _a,_b,_c;
_a = String.valueOf(a);
_b = String.valueOf(b);
_c = String.valueOf(c);
if(_b.length() == 1)
_b = "0"+_b;
if(_c.length() == 1)
_c = "0"+_c;
return _a+"-"+_b+"-"+_c;
}
public static boolean run(int a){
if((a%4==0 && a%100!=0) || a%400==0)
return true;
else
return false;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner reader = new Scanner(System.in);
String s = reader.next();
String _a,_b,_c;
_a = s.substring(0, 2);
_b = s.substring(3, 5);
_c = s.substring(6, 8);
int a,b,c;
a = Integer.parseInt(_a);
b = Integer.parseInt(_b);
c = Integer.parseInt(_c);
_a=rq(a,b,c);
_b=rq(c,a,b);
_c=rq(c,b,a);
TreeSet<String> ss = new TreeSet<String>();
if(_a.length()!=0) ss.add(_a);
if(_b.length()!=0) ss.add(_b);
if(_c.length()!=0) ss.add(_c);
for(String sss : ss){
System.out.println(sss);
}
}
}
PREV-40 k倍区间
简记：简单维护思想
#include
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
public class Main {
private static long ans = 0L;
private static int num;
public static void main(String[] args) throws IOException {
StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
in.nextToken();
int n = (int) in.nval;
in.nextToken();
int k = (int) in.nval;
ans = 0;
int[] sum = new int[100005];
int[] cnt = new int[100005];
cnt[0] = 1;
for (int i = 1; i <= n; i++) {
in.nextToken();
int t = (int) in.nval;
sum[i] = (sum[i - 1] + t) % k;
ans += cnt[sum[i]];
cnt[sum[i]]++;
}
out.println(ans);
out.flush();
}
}
PREV-41 Excel地址
简记：进制知识
#include import java.util.Scanner;
public class Main {
public static void main(String[] args){
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
char[] c = new char[100];
int cnt = 0;
while(n!=0){//500
if(n%26==0){
c[cnt++]=26+64;
n/=26;//19 // 0
n--;//18 //-1
}else{
c[cnt++]=(char) (n%26+64);//
n/=26;//n=1
}
}
for(int i=cnt-1;i>=0;i--)
System.out.print(c[i]);
}
}
PREV-42 九宫幻方
简记：dfs暴力搜索即可
#include PREV-43 拉马车
#include
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Stack;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String strA = br.readLine();
String strB = br.readLine();
br.close();
game(strA,strB);
}
private static void game(String strA, String strB) {
Stack<String> stack = new Stack<>();
while(true) {
while(true){
String a = strA.substring(0,1);
strA = strA.substring(1);
String temp = isWin(stack, a, strA);
if (temp.equals(strA)) {
break;
}else {
strA = temp;
}
}
if (strA.length() == 0) {
break;
}
while(true) {
String b = strB.substring(0, 1);
strB = strB.substring(1);
String temp = isWin(stack, b, strB);
if (temp.equals(strB)) {
break;
}else {
strB = temp;
}
}
if (strB.length() == 0) {
break;
}
}
if (strA.length() > strB.length()) {
System.out.println(strA);
}else {
System.out.println(strB);
}
}
private static String isWin(Stack<String> stack, String a, String str) {
if (stack.search(a) > 0) {
str += a;
while (true) {
String temp = (String)stack.pop();
str += temp;
if (temp.equals(a)) {
break;
}
}
}else {
stack.push(a);
}
return str;
}
}
PREV-44 青蛙跳杯子
简记：bfs
#include
import java.util.*;
public class Main {
static class Fron {
String now;
int step;
int pos;
Fron(String now, int step, int pos) {
this.now = now;
this.pos = pos;
this.step = step;
}
}
static String start, end;
static Queue<Fron> q = new ArrayDeque<Fron>();
static Map<String, Integer> m = new HashMap<String, Integer>();
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
start = sc.nextLine();
end = sc.nextLine();
int pos = 0;
for (int i = 0; i < start.length(); i++) {
if ('*' == start.charAt(i)) {
pos = i;
break;
}
}
System.out.println(bfs(start, 0, pos));
sc.close();
}
private static int bfs(String now, int step, int pos) {
Fron f = new Fron(now, step, pos);
q.offer(f);
while (!q.isEmpty()) {
Fron status = q.poll();
if (status.now.equals(end)) {
return status.step;
}
String temp;
if (m.containsKey(status.now)) {
continue;
} else {
m.put(status.now, 1);
}
if (status.pos + 1 < start.length()) {
temp = status.now;
temp = swap(status.pos, status.pos + 1, temp);
Fron fNext = new Fron(temp, status.step + 1, status.pos + 1);
if (!m.containsKey(temp)) {
q.offer(fNext);
}
}
if (status.pos + 2 < start.length()) {
temp = status.now;
temp = swap(status.pos, status.pos + 2, temp);
Fron fNext = new Fron(temp, status.step + 1, status.pos + 2);
if (!m.containsKey(temp)) {
q.offer(fNext);
}
}
if (status.pos + 3 < start.length()) {
temp = status.now;
temp = swap(status.pos, status.pos + 3, temp);
Fron fNext = new Fron(temp, status.step + 1, status.pos + 3);
if (!m.containsKey(temp)) {
q.offer(fNext);
}
}
if (status.pos - 1 > -1) {
temp = status.now;
temp = swap(status.pos, status.pos - 1, temp);
Fron fNext = new Fron(temp, status.step + 1, status.pos - 1);
if (!m.containsKey(temp)) {
q.offer(fNext);
}
}
if (status.pos - 2 > -1) {
temp = status.now;
temp = swap(status.pos, status.pos - 2, temp);
Fron fNext = new Fron(temp, status.step + 1, status.pos - 2);
if (!m.containsKey(temp)) {
q.offer(fNext);
}
}
if (status.pos - 3 > -1) {
temp = status.now;
temp = swap(status.pos, status.pos - 3, temp);
Fron fNext = new Fron(temp, status.step + 1, status.pos - 3);
if (!m.containsKey(temp)) {
q.offer(fNext);
}
}
}
return -1;
}
private static String swap(int i, int j, String now) {
char[] c = now.toCharArray();
char temp = c[i];
c[i] = c[j];
c[j] = temp;
now = new String(c);
return now;
}
}
PREV-45 图形排版
#include import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] arg) {
new Main().solve();
}
StringTokenizer ST;
BufferedReader BR;
PrintWriter PW;
String next() {
while(ST == null || !ST.hasMoreTokens()) {
try {
ST = new StringTokenizer(BR.readLine());
}catch (Exception e) {
// TODO: handle exception
throw new RuntimeException(e);
}
}
return ST.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
void solve() {
BR = new BufferedReader(new InputStreamReader(System.in));
PW = new PrintWriter(System.out);
int m = nextInt(), n = nextInt();
Pair a[] = new Pair[n + 10];
Triple cr[] = new Triple[n + 10];
cr[0] = new Triple();
//正向处理出加到第i块的状态，Triple记忆第i块右下坐标(x,y)和第i块缩放后的高度h
for(int i = 1; i <= n; i++) {
Triple tmp = new Triple(cr[i-1]);
if(tmp.x == m) tmp.x = tmp.h = 0;
a[i] = new Pair(nextInt(), nextInt());
cr[i] = new Triple();
Pair b = Change(a[i], m - tmp.x);
cr[i].x = tmp.x + b.x;
cr[i].h = Math.max(tmp.h, b.y);
cr[i].y = cr[i].h + tmp.y - tmp.h;
}
Triple A[] = new Triple[m];
Triple B[] = new Triple[m];
for(int i = 0; i < m; i++) {
A[i] = new Triple();
B[i] = new Triple();
}
int ans = cr[n].y;
for(int i = n; i >= 1; i--) {
//处理删除第i块的答案ah
Triple pre = cr[i-1];
int ah;
if(pre.x == m) {
ah = pre.y + B[0].y;
} else {
ah = pre.y - pre.h + B[pre.x].y - B[pre.x].h + Math.max(pre.h, B[pre.x].h);
}
ans = Math.min(ans, ah);
//逆向DP，处理出第i至n块从(0,j)位置开始放置
for(int j = 0; j < m; j++) {
Pair b = Change(a[i], m - j);
Triple tmp;
if(j + b.x == m) tmp = new Triple(0, B[0].y, 0);
else tmp = new Triple(B[j + b.x]);
A[j].h = Math.max(b.y, tmp.h);
A[j].y = A[j].h + tmp.y - tmp.h;
}
for(int j = 0; j < m; j++)
B[j] = new Triple(A[j]);
}
PW.print(ans);
PW.close();
}
Pair Change(Pair A, int x) {
if(A.x <= x) return new Pair(A);
return new Pair(x, (A.y * x + A.x - 1) / A.x);
}
}
class Pair implements Comparable<Pair> {
int x, y;
Pair() { }
Pair(Pair A) { x = A.x; y = A.y; }
Pair(int x, int y) {
this.x = x; this.y = y;
}
@Override
public int compareTo(Pair A) {
return x == A.x ? y - A.y : x - A.x;
}
}
class Triple {
int x, y, h;
Triple() {}
Triple(int x, int y, int h) {
this.x = x; this.y = y; this.h = h;
}
Triple(Triple A) {
x = A.x; y = A.y; h = A.h;
}
@Override
public String toString() {
return String.valueOf(x) + " " + String.valueOf(y) + " " + String.valueOf(h);
}
}