机考E卷100分题 - 喊7的次数重排

题目描述

喊7是一个传统的聚会游戏，N个人围成一圈，按顺时针从1到N编号。

编号为1的人从1开始喊数，下一个人喊的数字为上一个人的数字加1，但是当将要喊出来的数字是7的倍数或者数字本身含有7的话，不能把这个数字直接喊出来，而是要喊”过”。

假定玩这个游戏的N个人都没有失误地在正确的时机喊了”过”，当喊到数字K时，可以统计每个人喊”过”的次数。

现给定一个长度为N的数组，存储了打乱顺序的每个人喊”过”的次数，请把它还原成正确的顺序，即数组的第i个元素存储编号i的人喊”过”的次数。

输入描述

输入为一行，为空格分隔的喊”过”的次数，注意K并不提供，K不超过200，而数字的个数即为N。

输出描述

输出为一行，为顺序正确的喊”过”的次数，也由空格分隔。

示例1

输入

0 1 0
1

输出

1 0 0
1

说明

一共只有一次喊”过”，那只会发生在需要喊7时，按顺序，编号为1的人会遇到7，故输出1 0 0。

示例2

输入

一共只有一次喊”过”，那只会发生在需要喊7时，按顺序，编号为1的人会遇到7，故输出1 0 0。
1

输出

0 2 0 1 0
1

说明

一共有三次喊”过”，发生在7 14 17，按顺序，编号为2的人会遇到7 17，编号为4的人会遇到14，故输出0 2 0 1 0。

解题思路

Java

import java.util.Arrays;
import java.util.Scanner;
import java.util.StringJoiner;

public class Main {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);

    // 读取输入，将喊过的次数存储在数组中
    int[] counts =
        Arrays.stream(scanner.nextLine().split(" ")).mapToInt(Integer::parseInt).toArray();

    // 计算总共喊过的次数
    int totalPass = Arrays.stream(counts).sum();

    int numberOfPeople = counts.length;
    int[] peopleCounts = new int[numberOfPeople]; // 存储每个人喊过的次数

    int currentNumber = 1; // 当前要喊的数字
    int currentIndex = 0; // 当前喊过的人的索引
    while (totalPass > 0) {
      if (currentNumber % 7 == 0 || (currentNumber + "").contains("7")) { // 如果当前数字是7的倍数或包含7
        totalPass--;
        peopleCounts[currentIndex]++; // 当前喊过的人喊过的次数加1
      }
      currentNumber++;
      currentIndex = (currentIndex + 1) % numberOfPeople; // 更新喊过的人的索引
    }

    // 构造输出字符串
    StringJoiner stringJoiner = new StringJoiner(" ");
    for (int count : peopleCounts) {
      stringJoiner.add(Integer.toString(count));
    }
    System.out.println(stringJoiner.toString());
  }
}

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Python

counts = list(map(int, input().split()))

totalPass = sum(counts)

numberOfPeople = len(counts)
peopleCounts = [0] * numberOfPeople

currentNumber = 1
currentIndex = 0
while totalPass > 0:
    if currentNumber % 7 == 0 or str(currentNumber).find('7') != -1:
        totalPass -= 1
        peopleCounts[currentIndex] += 1
    currentNumber += 1
    currentIndex = (currentIndex + 1) % numberOfPeople

output = ' '.join(map(str, peopleCounts))
print(output)

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JavaScript

const readline = require('readline');

const rl = readline.createInterface({
  input: process.stdin,
  output: process.stdout
});

rl.on('line', (countsStr) => {
  let counts = countsStr.split(" ").map(Number);
  let totalPass = counts.reduce((sum, count) => sum + count, 0);
  let numberOfPeople = counts.length;
  let peopleCounts = new Array(numberOfPeople).fill(0);

  let currentNumber = 1;
  let currentIndex = 0;
  while (totalPass > 0) {
    if (currentNumber % 7 === 0 || currentNumber.toString().includes("7")) {
      totalPass--;
      peopleCounts[currentIndex]++;
    }
    currentNumber++;
    currentIndex = (currentIndex + 1) % numberOfPeople;
  }

  const result = peopleCounts.join(" ");
  console.log(result);

  rl.close();
});

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C++

#include <iostream>
#include <sstream>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
  string line;
  getline(cin, line);
  istringstream iss(line);
  
  vector<int> counts;
  int count;
  while (iss >> count) {
    counts.push_back(count);
  }
  
  int totalPass = 0;
  for (int i = 0; i < counts.size(); i++) {
    totalPass += counts[i];
  }
  
  int numberOfPeople = counts.size();
  vector<int> peopleCounts(numberOfPeople, 0);
  
  int currentNumber = 1;
  int currentIndex = 0;
  while (totalPass > 0) {
    if (currentNumber % 7 == 0 || to_string(currentNumber).find("7") != string::npos) {
      totalPass--;
      peopleCounts[currentIndex]++;
    }
    currentNumber++;
    currentIndex = (currentIndex + 1) % numberOfPeople;
  }
  
  string output;
  for (int i = 0; i < peopleCounts.size(); i++) {
    output += to_string(peopleCounts[i]) + " ";
  }
  output = output.substr(0, output.size() - 1);
  
  cout << output << endl;
  
  return 0;
}

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C语言

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

// 判断一个数字是否是7的倍数或包含数字7
int isPassNumber(int number) {
    if (number % 7 == 0) return 1;
    while (number > 0) {
        if (number % 10 == 7) return 1;
        number /= 10;
    }
    return 0;
}

int main() {
    char input[1000]; // 用于存储输入行的字符数组
    fgets(input, sizeof(input), stdin); // 读取输入的整行字符串

    // 计算有多少人（即数字个数）
    int numberOfPeople = 0;
    for (int i = 0; input[i] != '\0'; i++) {
        if (input[i] == ' ') numberOfPeople++;
    }
    numberOfPeople++; // 数字个数等于空格数加1

    int counts[numberOfPeople]; // 用于存储每个人的“喊过”次数
    int peopleCounts[numberOfPeople]; // 用于存储按顺序计算的每个人“喊过”次数
    memset(peopleCounts, 0, sizeof(peopleCounts)); // 初始化peopleCounts数组为0

    // 将输入的字符串转换为整数数组
    char* token = strtok(input, " ");
    for (int i = 0; i < numberOfPeople; i++) {
        counts[i] = atoi(token);
        token = strtok(NULL, " ");
    }

    // 计算总共的“喊过”次数
    int totalPass = 0;
    for (int i = 0; i < numberOfPeople; i++) {
        totalPass += counts[i];
    }

    int currentNumber = 1; // 当前要喊的数字
    int currentIndex = 0; // 当前要处理的人的索引

    // 模拟游戏过程
    while (totalPass > 0) {
        // 检查当前数字是否是7的倍数或包含7
        if (isPassNumber(currentNumber)) {
            totalPass--; // 总的“喊过”次数减少
            peopleCounts[currentIndex]++; // 当前人的“喊过”次数增加
        }
        currentNumber++; // 增加当前数字
        currentIndex = (currentIndex + 1) % numberOfPeople; // 更新当前人的索引
    }

    // 输出结果
    for (int i = 0; i < numberOfPeople; i++) {
        printf("%d", peopleCounts[i]);
        if (i != numberOfPeople - 1) {
            printf(" ");
        }
    }
    printf("\n");

    return 0;
}
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