(新卷,100分)- 开源项目热度榜单（Java JS Python C）

(新卷,100分)- 开源项目热度榜单（Java & JS & Python & C）

题目描述

某个开源社区希望将最近热度比较高的开源项目出一个榜单，推荐给社区里面的开发者。

对于每个开源项目，开发者可以进行关注（watch）、收藏（star）、fork、提issue、提交合并请求（MR）等。

数据库里面统计了每个开源项目关注、收藏、fork、issue、MR的数量，开源项目的热度根据这5个维度的加权求和进行排序。

H = W(watch) x #watch + W(star) x #star + W(fork) x #fork + W(issue) x #issue + W(mr) x #mr

• H 表示热度值
• W(watch)、W(star)、W(fork)、W(issue)、W(mr) 分别表示5个统计维度的权重
• #watch、#star、#fork、#issue、#mr 分别表示5个统计维度的统计值

榜单按照热度值降序排序，对于热度值相等的，按照项目名字转换为全小写字母后的字典序排序（'a','b','c',...,'x','y','z'）。

输入描述

第一行输入为N，表示开源项目的个数，0 ＜ N ＜100。

第二行输入为权重值列表，一共 5 个整型值，分别对应关注、收藏、fork、issue、MR的权重，权重取值 0 < W ≤ 50。

第三行开始接下来的 N 行为开源项目的统计维度，每一行的格式为：

name nr_watch nr_start nr_fork nr_issue nr_mr

其中 name 为开源项目的名字，由英文字母组成，长度 ≤ 50，其余 5 个整型值分别为该开源项目关注、收藏、fork、issue、MR的数量，数量取值 0 < nr ≤ 1000。

输出描述

按照热度降序，输出开源项目的名字，对于热度值相等的，按照项目名字转换为全小写后的字典序排序（'a' > 'b' > 'c' > ... > 'x' > 'y' > 'z'）。

用例

题目解析

简单的多条件排序题。

JavaScript算法源码

const rl = require("readline").createInterface({ input: process.stdin }); var iter = rl[Symbol.asyncIterator](); const readline = async () => (await iter.next()).value; void (async function () { class Project { constructor(name, hot) { this.name = name; this.hot = hot; } } const n = parseInt(await readline()); const weights = (await readline()).split(" ").map(Number); const projects = []; for (let i = 0; i < n; i++) { const tmp = (await readline()).split(" "); const name = tmp.shift(); let hot = 0; tmp.map(Number); for (let j = 0; j < 5; j++) { hot += tmp[j] * weights[j]; } projects.push(new Project(name, hot)); } projects .sort((a, b) => a.hot != b.hot ? b.hot - a.hot : strcmp(a.name.toLowerCase(), b.name.toLowerCase()) ) .forEach((p) => console.log(p.name)); })(); function strcmp(a, b) { if (a == b) { return 0; } else if (a > b) { return 1; } else { return -1; } }

Java算法源码

import java.util.Arrays; import java.util.Scanner; public class Main { static class Project { String name; int hot; public Project(String name, int hot) { this.name = name; this.hot = hot; } } public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); int[] weights = new int[5]; for (int i = 0; i < 5; i++) { weights[i] = sc.nextInt(); } Project[] projects = new Project[n]; for (int i = 0; i < n; i++) { String name = sc.next(); int hot = 0; for (int j = 0; j < 5; j++) { hot += sc.nextInt() * weights[j]; } projects[i] = new Project(name, hot); } Arrays.sort( projects, (a, b) -> a.hot != b.hot ? b.hot - a.hot : a.name.toLowerCase().compareTo(b.name.toLowerCase())); for (Project project : projects) { System.out.println(project.name); } } }

Python算法源码

class Project: def __init__(self, name, hot): self.name = name self.hot = hot # 输入获取 n = int(input()) weights = list(map(int, input().split())) projects = [] for _ in range(n): tmp = input().split() name = tmp.pop(0) statistics = list(map(int, tmp)) hot = 0 for i in range(5): hot += statistics[i] * weights[i] projects.append(Project(name, hot)) projects.sort(key=lambda x: (-x.hot, x.name.lower())) for p in projects: print(p.name)

C算法源码

#include <stdio.h> #include <stdlib.h> #include <string.h> #define MAX_NAME_LEN 51 typedef struct { char name[MAX_NAME_LEN]; int hot; } Project; Project *new_Project(char *name, int hot) { Project *p = (Project *) malloc(sizeof(Project)); strcpy(p->name, name); p->hot = hot; return p; } char *lower(char *s) { char *res = (char *) calloc(MAX_NAME_LEN, sizeof(char)); strcpy(res, s); int diff = 'a' - 'A'; int i = 0; while (res[i] != '\0') { if (res[i] >= 'A' && res[i] <= 'Z') { res[i] = (char) (res[i] + diff); } i++; } return res; } int cmp(const void *a, const void *b) { Project *A = *((Project **) a); Project *B = *((Project **) b); if (A->hot != B->hot) { return B->hot - A->hot; } else { return strcmp(lower(A->name), lower(B->name)); } } int main() { int n; scanf("%d", &n); int weights[5]; for (int i = 0; i < 5; i++) { scanf("%d", &weights[i]); } Project *ps[n]; for (int i = 0; i < n; i++) { char name[MAX_NAME_LEN]; scanf("%s", name); int hot = 0; for (int j = 0; j < 5; j++) { int val; scanf("%d", &val); hot += val * weights[j]; } ps[i] = new_Project(name, hot); } qsort(ps, n, sizeof(Project *), cmp); for (int i = 0; i < n; i++) { puts(ps[i]->name); } return 0; }