第十四届蓝桥杯大赛软件赛省赛C/C++ 大学 B 组
A题：日期统计
直接遍历2023年每一天，看数组中是否有符合的
java的coding如下：
java
public class Main {
public static void main(String[] args) {
int[] array = {
5, 6, 8, 6, 9, 1, 6, 1, 2, 4, 9, 1, 9, 8, 2, 3, 6, 4, 7, 7,
5, 9, 5, 0, 3, 8, 7, 5, 8, 1, 5, 8, 6, 1, 8, 3, 0, 3, 7, 9,
2, 7, 0, 5, 8, 8, 5, 7, 0, 9, 9, 1, 9, 4, 4, 6, 8, 6, 3, 3,
8, 5, 1, 6, 3, 4, 6, 7, 0, 7, 8, 2, 7, 6, 8, 9, 5, 6, 5, 6,
1, 4, 0, 1, 0, 0, 9, 4, 8, 0, 9, 1, 2, 8, 5, 0, 2, 5, 3, 3
};
int[] daysInMonth = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int ans = 0;
for (int month = 1; month <= 12; ++month) {
for (int day = 1; day <= daysInMonth[month]; ++day) {
int[] dateSeq = {2, 0, 2, 3, month / 10, month % 10, day / 10, day % 10};
int k = 0;
for (int i = 0; i < 100; ++i) {
if (array[i] == dateSeq[k]) {
++k;
if (k == 8) {
ans++;
break;
}
}
}
}
}
System.out.println(ans);
}
}go的coding如下：
go
package main
import "fmt"
func main() {
array := []int{
5, 6, 8, 6, 9, 1, 6, 1, 2, 4, 9, 1, 9, 8, 2, 3, 6, 4, 7, 7,
5, 9, 5, 0, 3, 8, 7, 5, 8, 1, 5, 8, 6, 1, 8, 3, 0, 3, 7, 9,
2, 7, 0, 5, 8, 8, 5, 7, 0, 9, 9, 1, 9, 4, 4, 6, 8, 6, 3, 3,
8, 5, 1, 6, 3, 4, 6, 7, 0, 7, 8, 2, 7, 6, 8, 9, 5, 6, 5, 6,
1, 4, 0, 1, 0, 0, 9, 4, 8, 0, 9, 1, 2, 8, 5, 0, 2, 5, 3, 3,
}
daysInMonth := []int{0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}
ans := 0
for month := 1; month <= 12; month++ {
for day := 1; day <= daysInMonth[month]; day++ {
dateSeq := [8]int{2, 0, 2, 3, month / 10, month % 10, day / 10, day % 10}
k := 0
for i := 0; i < 100; i++ {
if array[i] == dateSeq[k] {
k++
if k == 8 {
ans++
break
}
}
}
}
}
fmt.Println(ans)
}B题：01串的熵
枚举即可
注意设置变量为浮点型，避免多余的类型转换
java的coding如下：
java
import java.util.Scanner;
// 1:无需package
// 2: 类名必须Main, 不可修改
public class Main {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
//在此输入您的代码...
double n = 23333333;
for (int i = 0; i <= n / 2; i++) {
double sum = 0;
sum -= i * (i / n) * Math.log(i / n) / Math.log(2) + (n - i) * ((n - i) / n) * Math.log((n - i) / n) / Math.log(2);
if (sum > 11625907.5 && sum < 11625907.6) {
System.out.println(i);
break;
}
}
scan.close();
}
}go的coding如下：
go
package main
import (
"fmt"
"math"
)
func main() {
n := 23333333.0
var sum float64
for o := 0; o <= int(n/2); o++ {
sum = 0
sum -= float64(o) * (float64(o) / n) * math.Log(float64(o)/n)/math.Log(2) + (n - float64(o)) * ((n - float64(o)) / n) * math.Log((n - float64(o))/n)/math.Log(2)
if sum > 11625907.5 && sum < 11625907.6 {
fmt.Println(o)
break
}
}
}C题：冶炼金属
推公式题目
java的coding如下：
java
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int n = scan.nextInt();
int[] a = new int[n];
int[] b = new int[n];
for (int i = 0; i < n; i++) {
int A = scan.nextInt();
int B = scan.nextInt();
a[i] = A / B;
b[i] = A / (B + 1);
}
int maxB = b[0];
int minA = a[0];
for (int i = 1; i < n; i++) {
maxB = Math.max(maxB, b[i]);
minA = Math.min(minA, a[i]);
}
System.out.println((maxB + 1) + " " + minA);
scan.close();
}
}go的coding如下：
go
package main
import (
"fmt"
)
func main() {
var n int
fmt.Scan(&n)
a := make([]int, n)
b := make([]int, n)
for i := 0; i < n; i++ {
var A, B int
fmt.Scan(&A, &B)
a[i] = A / B
b[i] = A / (B + 1)
}
maxB := b[0]
minA := a[0]
for i := 1; i < n; i++ {
if b[i] > maxB {
maxB = b[i]
}
if a[i] < minA {
minA = a[i]
}
}
fmt.Println(maxB+1, minA)
}D题：飞机降落
DFS暴力搜索即可 记得回溯
java的coding如下：
java
import java.util.Scanner;
public class Main {
static class Plane {
int t, d, l;
}
static boolean[] vis = new boolean[15];
static boolean flag;
static Plane[] p = new Plane[15];
public static void dfs(int m, int cnt, int last) {
if (cnt == m) {
flag = true;
return;
}
for (int i = 0; i < m; i++) {
if (!vis[i] && p[i].t + p[i].d >= last) {
vis[i] = true;
dfs(m, cnt + 1, Math.max(last, p[i].t) + p[i].l);
vis[i] = false;
}
}
}
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int T = scan.nextInt();
for (int t = 0; t < T; t++) {
int N = scan.nextInt();
for (int i = 0; i < N; i++) {
p[i] = new Plane();
p[i].t = scan.nextInt();
p[i].d = scan.nextInt();
p[i].l = scan.nextInt();
}
flag = false;
dfs(N, 0, 0);
if (flag) {
System.out.println("YES");
} else {
System.out.println("NO");
}
}
scan.close();
}
}go的coding如下：
go
package main
import "fmt"
type plane struct {
t, d, l int
}
var vis [15]bool
var flag bool
var p [15]plane
func dfs(m, cnt, last int) {
if cnt == m {
flag = true
return
}
for i := 0; i < m; i++ {
if !vis[i] && p[i].t+p[i].d >= last {
vis[i] = true
dfs(m, cnt+1, max(last, p[i].t)+p[i].l)
vis[i] = false
}
}
}
func max(a, b int) int {
if a > b {
return a
}
return b
}
func main() {
var T, N int
fmt.Scan(&T)
for t := 0; t < T; t++ {
fmt.Scan(&N)
for i := 0; i < N; i++ {
fmt.Scan(&p[i].t, &p[i].d, &p[i].l)
}
flag = false
dfs(N, 0, 0)
if flag {
fmt.Println("YES")
} else {
fmt.Println("NO")
}
}
}E题：接龙数列
咳咳，这边暴力只通过40%
java
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int n = scan.nextInt();
List<String> list = new ArrayList<>();
for (int i = 0; i < n; i++) {
list.add(scan.next());
}
scan.close();
int res = 0;
Iterator<String> it = list.iterator();
String prev = it.next();
while (it.hasNext()) {
String current = it.next();
if (prev.charAt(prev.length() - 1) != current.charAt(0)) {
it.remove();
res++;
n--;
} else {
prev = current;
}
}
System.out.println(res);
}
}所以需要使用动态规划
根据数列总长度-最长接龙数列长度等于最少删除次数求最优解， 定义状态为当前最长接龙数列长度，则dp[i]就是以i为数字最后一位的最长接龙数列长度，设x为当前数字的第一位(如果为接龙数列，也就是前一位数的最后一位)，y为当前数字的最后一位，则转移方程可以写为dp[y]=max(dp[x]+1,dp[y])
java的coding如下：
java
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int n = scan.nextInt();
int[] dp = new int[10];
int m = 0;
for (int i = 0; i < n; i++) {
String s = scan.next();
int x = s.charAt(0) - '0';
int y = s.charAt(s.length() - 1) - '0';
dp[y] = Math.max(dp[x] + 1, dp[y]); //选与不选 不选即是删除
m = Math.max(m, dp[y]);
}
System.out.println(n - m);
scan.close();
}
}go的coding如下：
go
package main
import (
"fmt"
)
func main() {
var n, x, y, m int
fmt.Scan(&n)
dp := make([]int, 10)
for i := 0; i < n; i++ {
var s string
fmt.Scan(&s)
x = int(s[0] - '0')
y = int(s[len(s)-1] - '0')
dp[y] = max(dp[x]+1, dp[y])
m = max(m, dp[y])
}
fmt.Println(n - m)
}
func max(a, b int) int {if a > b { return a }; return b}