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腾讯云wordpress搭建网站,江西网站开发费用,网页版梦幻西游勇闯火焰山攻略,品牌设计工作内容(新卷,100分)- 计算最大乘积#xff08;Java JS Python C#xff09;题目描述给定一个元素类型为小写字符串的数组#xff0c;请计算两个没有相同字符的元素长度乘积的最大值#xff0c;如果没有符合条件的两个元素#xff0c;返回0。输入描述输入为一个…(新卷,100分)- 计算最大乘积Java JS Python C题目描述给定一个元素类型为小写字符串的数组请计算两个没有相同字符的元素长度乘积的最大值如果没有符合条件的两个元素返回0。输入描述输入为一个半角逗号分隔的小写字符串的数组2 数组长度1000 字符串长度 50。输出描述两个没有相同字符的元素 长度乘积的最大值。用例输入iwdvpbn,hk,iuop,iikd,kadgpf输出14说明数组中有5个元素。iwdvpbn与hk无相同的字符满足条件iwdvpbn的长度为7hk的长度为2乘积为147*2。iwdvpbn与iuop、iikd、kadgpf均有相同的字符不满足条件。iuop与iikd、kadgpf均有相同的字符不满足条件。iikd与kadgpf有相同的字符不满足条件。因此输出为14。位运算解法JS算法源码/* JavaScript Node ACM模式 控制台输入获取 */ const readline require(readline); const rl readline.createInterface({ input: process.stdin, output: process.stdout, }); rl.on(line, (line) { console.log(getResult(line.split(,))); }); function getResult(words) { let ans 0; const n words.length; const bits new Array(n).fill(0); for (let i 0; i n; i) { for (let j 0; j words[i].length; j) { bits[i] | 1 (words[i][j].charCodeAt() - 97); } } for (let i 0; i n; i) { for (let j i 1; j n; j) { if ((bits[i] bits[j]) 0) { ans Math.max(ans, words[i].length * words[j].length); } } } return ans; }Java算法源码import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc new Scanner(System.in); System.out.println(getResult(sc.nextLine().split(,))); } public static int getResult(String[] words) { int ans 0; int n words.length; int[] bits new int[n]; for (int i 0; i n; i) { for (int j 0; j words[i].length(); j) { bits[i] | 1 (words[i].charAt(j) - a); } } for (int i 0; i n; i) { for (int j i 1; j n; j) { if ((bits[i] bits[j]) 0) { ans Math.max(ans, words[i].length() * words[j].length()); } } } return ans; } }Python算法源码# 输入获取 words input().split(,) # 算法入口 def getResult(): ans 0 n len(words) bits [0] * n for i in range(n): for j in range(len(words[i])): bits[i] | (1 (ord(words[i][j]) - 97)) for i in range(n): for j in range(i 1, n): if (bits[i] bits[j]) 0: ans max(ans, len(words[i]) * len(words[j])) return ans # 算法调用 print(getResult())C算法源码#include stdio.h #include stdlib.h #include string.h #define MAX(a,b) (a) (b) ? (a) : (b) int getResult(char ** words, int wordsSize); int main() { char s[5000]; gets(s); char* words[100]; int wordsSize 0; char* token strtok(s, ,); while(token ! NULL) { words[wordsSize] token; token strtok(NULL, ,); } printf(%d\n, getResult(words, wordsSize)); return 0; } int getResult(char ** words, int wordsSize) { int ans 0; int* bits (int*) calloc(wordsSize, sizeof(int)); for(int i0; iwordsSize; i) { for(int j0; jstrlen(words[i]); j) { bits[i] | 1 (words[i][j] - a); } } for(int i0; iwordsSize; i) { for(int ji1; jwordsSize; j) { if((bits[i] bits[j]) 0) { ans MAX(ans, strlen(words[i]) * strlen(words[j])); } } } return ans; }交集解法JavaScript算法源码/* JavaScript Node ACM模式 控制台输入获取 */ const readline require(readline); const rl readline.createInterface({ input: process.stdin, output: process.stdout, }); rl.on(line, (line) { const arr line.split(,); console.log(getResult(arr)); }); function getResult(arr) { let ans 0; const sets arr.map((s) new Set([...s])); for (let i 0; i sets.length; i) { for (let j i 1; j sets.length; j) { if (disjoint(sets[i], sets[j])) { ans Math.max(ans, arr[i].length * arr[j].length); } } } return ans; } function disjoint(set1, set2) { for (let c of set1) { if (set2.has(c)) return false; } return true; }Java算法源码import java.util.ArrayList; import java.util.Collections; import java.util.HashSet; import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc new Scanner(System.in); String[] strings sc.nextLine().split(,); System.out.println(getResult(strings)); } public static int getResult(String[] arr) { ArrayListHashSetCharacter list new ArrayList(); for (String s : arr) { HashSetCharacter set new HashSet(); for (char c : s.toCharArray()) set.add(c); list.add(set); } int ans 0; for (int i 0; i list.size(); i) { HashSetCharacter a list.get(i); for (int j i 1; j list.size(); j) { HashSetCharacter b list.get(j); if (Collections.disjoint(a, b)) { ans Math.max(ans, arr[i].length() * arr[j].length()); } } } return ans; } }Python算法源码# 输入获取 arr input().split(,) # 算法入口 def getResult(): sets list(map(lambda x: set(x), arr)) ans 0 for i in range(len(sets)): for j in range(i1, len(sets)): if sets[i].isdisjoint(sets[j]): ans max(ans, len(arr[i]) * len(arr[j])) return ans # 算法调用 print(getResult())