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第十二届蓝桥杯决赛JavaC组真题——详细答案对照(全网唯一:异或变换100%数据)

发布时间:2024/8/26 java 62 豆豆
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目录

A、整数范围

B 、带宽

C、纯质数

D 、完全日期

E、最小权值

F、大写

G、123

H、异或变换(真真的搞不出来)

I、冰山

J、二进制问题


A、整数范围

本题总分:5 分

问题描述

  用 8 位二进制(一个字节)来表示一个非负整数,表示的最小值是 0 ,则一般能表示的最大值是多少?

答案提交

  这是一道结果填空的题,你只需要算出结果后提交即可。本题的结果为一个整数,在提交答案时只填写这个整数,填写多余的内容将无法得分。
 

package action;public class demo {public static void main(String[] args) {System.out.println(Integer.parseInt("11111111", 2));} }

B 带宽

本题总分:5 分

问题描述

  小蓝家的网络带宽是 200 Mbps,请问,使用小蓝家的网络理论上每秒钟最多可以从网上下载多少 MB 的内容。

答案提交

  这是一道结果填空的题,你只需要算出结果后提交即可。本题的结果为一个整数,在提交答案时只填写这个整数,填写多余的内容将无法得分。

package action;public class demo {public static void main(String[] args) {System.out.println(200/8);} }

C、纯质数

本题总分:10 分

问题描述

  如果一个正整数只有 1和它本身两个约数,则称为一个质数(又称素数)。
  前几个质数是:2,3,5,7,11,13,17,19,23,29,31,37,⋅⋅⋅ 。
  如果一个质数的所有十进制数位都是质数,我们称它为纯质数。例如:2,3,5,7,23,37 都是纯质数,而 11,13,17,19,29,31 不是纯质数。当然1,4,35 也不是纯质数。
  请问,在 1 到 20210605 中,有多少个纯质数?

答案提交

  这是一道结果填空的题,你只需要算出结果后提交即可。本题的结果为一个整数,在提交答案时只填写这个整数,填写多余的内容将无法得分。

package action;public class demo {public static void main(String[] args) {long out = 0;// 记录总个数for (int i = 1; i <= 20210605; i++) {if ((i + "").indexOf("1") == -1 && (i + "").indexOf("4") == -1 && (i + "").indexOf("6") == -1 && // 首先判断当前数字中不能包含非质数数字(i + "").indexOf("8") == -1 && (i + "").indexOf("9") == -1 && (i + "").indexOf("0") == -1) {if (f(i)) { // 判断是否是质数out++;// System.out.println(i);}}if (i % 10000 == 0) {System.out.println(i);// 观察程序运行进度}}System.out.println(out);// 最后结果}// 判断是否是质数private static boolean f(int num) {for (int i = 2; i < num / 2; i++) {if (num % i == 0) {return false;}}return true;} }

D 完全日期

本题总分:10 分

问题描述

  如果一个日期中年月日的各位数字之和是完全平方数,则称为一个完全日期。
  例如:2021 年 6 月 5 日的各位数字之和为 2+0+2+1+6+5=16,而 16 是一个完全平方数,它是 4 的平方。所以 2021 年 6 月 5 日是一个完全日期。
  例如:2021 年 6 月 23 日的各位数字之和为 2+0+2+1+6+2+3=16,是一个完全平方数。所以 2021 年 6 月 23 日也是一个完全日期。
  请问,从 2001 年 1 月 1 日到 2021 年 12 月 31 日中,一共有多少个完全日期?

答案提交

  这是一道结果填空的题,你只需要算出结果后提交即可。本题的结果为一个整数,在提交答案时只填写这个整数,填写多余的内容将无法得分。

package action;import java.util.Calendar;public class demo {public static void main(String[] args) {long out = 0;// 记录总个数Calendar n = Calendar.getInstance();n.clear();for (int i = 2001; i < 2022; i++) {// 年份for (int j = 1; j <= 12; j++) {// 月份n.set(i, j - 1, 1);int max = n.getActualMaximum(Calendar.DAY_OF_MONTH);// 获取当前月份最大天数for (int k = 1; k <= max; k++) {// 日// System.out.println(i+""+j+""+k);if (f(Integer.parseInt(i + "" + j + "" + k))) {out++;}}}}System.out.println(out);// 最后结果}// 判断是否是完全日期private static boolean f(int num) {int n = 0;while (num > 0) {n += num % 10;num /= 10;}if (n == 4 || n == 9 || n == 16 || n == 25) {// 满足当前范围内日期的完全平方数只有4 9 16 25无需判断其余数字return true;}return false;} }

E、最小权值

本题总分:15 分

问题描述

  对于一棵有根二叉树 T ,小蓝定义这棵树中结点的权值 W(T) 如下:
  空子树的权值为 0 。
  如果一个结点 v 有左子树 L , 右子树 R ,分别有 C(L) 和 C(R) 个结点,则 W(v)=1+2W(L)+3W(R)+(C(L))^2C(R)。(^2代表平方)
  树的权值定义为树的根结点的权值。
  小蓝想知道,对于一棵有 2021 个结点的二叉树,树的权值最小可能是多少?

答案提交

  这是一道结果填空的题,你只需要算出结果后提交即可。本题的结果为一个整数,在提交答案时只填写这个整数,填写多余的内容将无法得分。

package action;public class demo {static int n = 2021;public static void main(String[] args) {f();}public static void f() {long[] dp = new long[n + 1];dp[1] = 1;for (int i = 2; i <= n; ++i) {long min = Long.MAX_VALUE;for (int l = 0; l <= i - 1; ++l) {int r = i - l - 1;long p = 1 + 2 * dp[l] + 3 * dp[r] + l * l * r;min = Math.min(min, p);}dp[i] = min;}System.out.println(dp[n]);}}

F、大写

时间限制: 1.0s 内存限制: 512.0MB 本题总分:15 分

问题描述

  给定一个只包含大写字母和小写字母的字符串,请将其中所有的小写字母转换成大写字母后将字符串输出。

输入格式

  输入一行包含一个字符串。

输出格式

  输出转换成大写后的字符串。
测试样例1
Input:
LanQiao

Output:
LANQIAO
评测用例规模与约定

  对于所有评测用例,字符串的长度不超过100。

package action;import java.util.Scanner;public class demo {public static void main(String[] args) {Scanner sc = new Scanner(System.in);System.out.println(sc.next().toUpperCase());}}

G、123

时间限制: 5.0s 内存限制: 512.0MB 本题总分:20 分

问题描述

  小蓝发现了一个有趣的数列,这个数列的前几项如下:
  1,1,2,1,2,3,1,2,3,4,...
  小蓝发现,这个数列前 1 项是整数 1 ,接下来 2 项是整数 1 至 2 接下来 3 项是整数 1 至 3 接下来 4 项是整数 1 至 4 ,依次类推。
  小蓝想知道,这个数列中,连续一段的和是多少。

输入格式

  输入的第一行包含一个整数 T ,表示询问的个数。
  接下来 T 行,每行包含一组询问,其中第 i 行包含两个整数 li和 ri,表示询问数列中第 li个数到第 ri个数的和。

输出格式

  输出 T 行,每行包含一个整数表示对应询问的答案。

测试样例1
Input:
3
1 1
1 3
5 8

Output:
1
4
8
评测用例规模与约定

  对于 10 1010% 的评测用例,1 ≤ T ≤ 30 , 1 ≤ l i ≤ r i ≤ 100。
  对于 20 2020% 的评测用例,1 ≤ T ≤ 100 , 1 ≤ l i ≤ r i ≤ 1000。
  对于 40 4040% 的评测用例,1 ≤ T ≤ 1000 , 1 ≤ l i ≤ r i ≤ 10^6 。
  对于 70 7070% 的评测用例,1 ≤ T ≤ 10000 , 1 ≤ l i ≤ r i ≤ 10^9。
  对于 80 8080% 的评测用例,1 ≤ T ≤ 1000 , 1 ≤ l i ≤ r i ≤ 10^12。
  对于 90 9090% 的评测用例,1 ≤ T ≤ 10000 , 1 ≤ l i ≤ r i ≤ 10^12。
  对于所有评测用例,1 ≤ T ≤ 100000 , 1 ≤ l i ≤ r i ≤ 1 0^12。

package action;import java.util.Scanner;public class demo {public static void main(String[] args) {Scanner sc = new Scanner(System.in);int t = sc.nextInt();for (int i = 0; i < t; ++i) {long l = sc.nextLong(), r = sc.nextLong();long ans_l = ffa(l - 1), ans_r = ffa(r);System.out.println(ans_r - ans_l);}}public static long ffa(long r) {long ind = 1, cnt = 1, ans = 0;while (ind <= r) {ans += cnt * (cnt + 1) / 2;++cnt;ind += cnt;}if (ind > r) {ind -= cnt;cnt = r - ind;ans += cnt * (cnt + 1) / 2;}return ans;}}

H、异或变换(源码提供者:学生【郭尚】)

时间限制: 3.0s 内存限制: 512.0MB 本题总分:20 分

问题描述

  小蓝有一个 01 串 s = s1s2s3 ⋅ ⋅ ⋅ sn。
  以后每个时刻,小蓝要对这个 01 串进行一次变换。每次变换的规则相同。
  对于 01 串 s = s1s2s3 ⋅ ⋅ ⋅ sn,变换后的 01 串s' = s'1s'2s'3 ⋅ ⋅ ⋅ s'n为:
  s'1=s1;    
  s'i=si-1⊕si。
  其中 a ⊕ b 表示两个二进制的异或,当 a 和 b 相同时结果为 0 ,当 a 和 b
  不同时结果为 1。
  请问,经过 t 次变换后的 01 串是什么?

输入格式

  输入的第一行包含两个整数 n,t,分别表示 01 串的长度和变换的次数。
  第二行包含一个长度为 n 的 01 串。

输出格式

  输出一行包含一个 01 串,为变换后的串。

测试样例1
Input:
5 3
10110

Output:
11010

Explanation:
初始时为 10110,变换 1 次后变为 11101,变换 2 次后变为 10011,变换 3 次后变为 11010。

评测用例规模与约定

  对于 40% 的评测用例,1 ≤ n ≤ 100 , 1 ≤ t ≤ 1000。
  对于 80% 的评测用例,1 ≤ n ≤ 1000 , 1 ≤ t ≤ 10^9。
  对于所有评测用例,1 ≤ n ≤ 10000 , 1 ≤ t ≤ 10^18。

package action;import java.util.Scanner;public class demo {public static void main(String[] args) {Scanner sc = new Scanner(System.in);int n = sc.nextInt();long t = sc.nextInt();// 变换次数StringBuilder as = new StringBuilder();// 输出得变量char[] c = sc.next().toCharArray();// 一个长度为n得字符串转数组sc.close();for (int i = 0; i < n; i++) {boolean a = c[i] == '0';// true是0// 在n里面判断是否相等for (int j = 1; i - j >= 0 && j <= t; j++)if ((t & j) == j && c[i - j] == '1')a = !a;as.append(a ? 0 : 1);// 三目运算,相等是0,不相等是1}System.out.println(as);} }

测试数据:1 ≤ n ≤ 10000 , 1 ≤ t ≤ 10^18

1000 100000000 10110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110010111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111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

结果:

1000 100000000
10110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010110101101011010011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000110001100011000011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110011100111001110010111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111011110111101111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I、冰山

时间限制: 5.0s 内存限制: 512.0MB 本题总分:25 分

问题描述

  一片海域上有一些冰山,第 i ii 座冰山的体积为 Vi。
  随着气温的变化,冰山的体积可能增大或缩小。第 i 天,每座冰山的变化量都是 Xi。当 Xi > 0 时,所有冰山体积增加 Xi;当 Xi < 0 时,所有冰山体积减少 −Xi ;当 Xi = 0 时,所有冰山体积不。
  如果第 i 天某座冰山的体积变化后小于等于 0 ,则冰山会永远消失。
  冰山有大小限制 k 。如果第 i 天某座冰山 j 的体积变化后 Vj大于 k ,则它会分裂成一个体积为 k 的冰山和 Vj − k座体积为 1 的冰山。
  第 i 天结束前(冰山增大、缩小、消失、分裂完成后),会漂来一座体积为 Yi的冰山(Yi = 0 表示没有冰山漂来)。
  小蓝在连续的 m mm 天对这片海域进行了观察,并准确记录了冰山的变化。小蓝想知道,每天结束时所有冰山的体积之和(包括新漂来的)是多少。
  由于答案可能很大,请输出答案除以 998244353 的余数。

输入格式

  输入的第一行包含三个整数 n,m,k,分别表示初始时冰山的数量、观察的天数以及冰山的大小限制。
  第二行包含 n nn 个整数 V1, V2 , ⋅⋅⋅ , Vn,表示初始时每座冰山的体积。
  接下来 m 行描述观察的 m 天的冰山变化。其中第 i 行包含两个整数 Xi , Yi,意义如前所述。

输出格式

  输出 m 行,每行包含一个整数,分别对应每天结束时所有冰山的体积之和除以 998244353 的余数。

测试样例1
Input:
1 3 6
1
6 1
2 2
-1 1

Output:
8
16
11

Explanation:
在本样例说明中,用 [a1, a2, · · · , an] 来表示每座冰山的体积。
初始时的冰山为 [1]。
第 1 天结束时,有 3 座冰山:[1, 1, 6]。
第 2 天结束时,有 6 座冰山:[1, 1, 2, 3, 3, 6]。
第 3 天结束时,有 5 座冰山:[1, 1, 2, 2, 5]。
评测用例规模与约定

  对于 40 4040% 的评测用例,n,m,k≤2000;
  对于 60 6060% 的评测用例,n,m,k≤20000;
  对于所有评测用例,1 ≤ n , m ≤ 100000 , 1 ≤ k ≤ 10^9 , 1 ≤ Vi ≤ k , 0 ≤ Yi ≤ k , − k ≤ Xi ≤ k。

package action;import java.util.LinkedList; import java.util.Queue; import java.util.Scanner;public class demo {public static void main(String[] args) {Scanner sc = new Scanner(System.in);int n = sc.nextInt();int m = sc.nextInt();long k = sc.nextInt();Queue<Long> que = new LinkedList<Long>();for (int i = 0; i < n; i++) {// 初始时冰山状态que.add(sc.nextLong());}for (int i = 0; i < m; i++) {//long x = sc.nextLong();long y = sc.nextLong();long sum = y;if (x != 0) {int len = que.size();// 队列循环时长度会变化 单独记录一下for (int j = 0; j < len; j++) {long temp = que.poll() + x;// 计算当前冰山状态 大于0再加入回去if (temp > 0) {sum += temp;if (temp > k) {que.add(k);// 体积大于k添加一个k 其余为1for (int l = 0; l < temp - k; l++) {que.add(1l);}} else {que.add(temp);}}}}if (y != 0) {// 判断有没有新冰山que.add(y);}System.out.println(sum % 998244353l);}} }

J、二进制问题

时间限制: 1.0s 内存限制: 512.0MB 本题总分:25 分

问题描述

  小蓝最近在学习二进制。他想知道 1 到 N 中有多少个数满足其二进制表示中恰好有 K 个 1。你能帮助他吗?

输入格式

  输入一行包含两个整数 N 和 K。

输出格式

  输出一个整数表示答案。

测试样例1
Input:
7 2

Output:
3
评测用例规模与约定

  对于 30 3030% 的评测用例,1 ≤ N ≤ 10^6 , 1 ≤ K ≤ 10。
  对于 60 6060% 的评测用例,1 ≤ N ≤ 2 × 10^9 , 1 ≤ K ≤ 30。
  对于所有评测用例,1 ≤ N ≤ 1 0^18 , 1 ≤ K ≤ 50。

package action;import java.util.Scanner;public class demo {public static long out = 0;public static int[] aa;public static long n;public static int k;public static void main(String[] args) {Scanner sc = new Scanner(System.in);n = sc.nextLong();k = sc.nextInt();sc.close();aa = new int[Long.toString(n, 2).length()];f(k, 0);System.out.println(out);}private static void f(int kk, int index) {if (kk == 0) {StringBuilder sb = new StringBuilder();for (int i = 0; i < aa.length; i++) {sb.append(aa[i]);}if (Long.valueOf(sb.toString(), 2) <= n) {out++;}} else {if (aa.length - index < kk) {return;} else {aa[index] = 1;f(kk - 1, index + 1);aa[index] = 0;f(kk, index + 1);}}} }

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