CodinGame: Temperatures 反思

Temperatures 温度

WHAT WILL I LEARN? 我能学到什么？
Conditions,Loops,Arrays 选择结构(if 语句)、循环结构(for/ while 语句)、数组

-Solving this puzzle validates(确认) that the loop concept is understood and that you can compare a list of values.
（解决这个难题可以验证你对循环的概念是否理解，你可以通过比较一个列表的变量/值）
-This puzzle is also a playground to experiment the concept of lambdas(λ, 匿名函数) in different programming languages. It’s also an opportunity to discover functional programming.
（它是一个用不同编程语言的实验的游乐场，也是一个发现函数编程的机会）

---

The Goal
In this exercise, you have to analyze records of temperature to find the closest to zero.

Rules
Write a program that prints the temperature closest to 0 among input data. If two numbers are equally close to zero, positive integer has to be considered closest to zero (for instance, if the temperatures are -5 and 5, then display 5).

原程序

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>

using namespace std;

/**
 * Auto-generated code below aims at helping you parse
 * the standard input according to the problem statement.
 **/
int main()
{
    int n; // the number of temperatures to analyse
    cin >> n; cin.ignore();
    string temps; // the n temperatures expressed as integers ranging from -273 to 5526
    getline(cin, temps);

    // Write an action using cout. DON'T FORGET THE "<< endl"
    // To debug: cerr << "Debug messages..." << endl;

    cout << "result" << endl;
}

关卡

01 Simple test case 简单的考验

Standard Output Stream:

Input messages are n = 5 temps = 1 -2 -8 4 5
output 1
有n个数，数列为 temps (string)

02 Only negative numbers 只有负数（零下）

Standard Output Stream:

Input messages are n = 3 temps = -12 -5 -137
output -5

03 Choose the right temperature 选出正确的温度

Standard Output Stream:

Input messages are n = 6 temps = 42 -5 12 21 5 24
output 5

04 Choose the right temperature 2 选出正确的温度2

Standard Output Stream:

Input messages are n = 6 temps = 42 5 12 21 -5 24
output 5

05 Complex test case 复杂的考验

Standard Output Stream:

Input messages are n = 10 temps = -5 -4 -2 12 -40 4 2 18 11 5
output 2

06 No temperature 没有温度

Standard Output Stream:

Input messages are n = 0 temps = (none)
output 0

过关思路

很明显，这是一个比大小的问题，要求的是：数列中绝对值最小的数
难点就在于：如何将string数组（字符串数组）转化为int数组（整型数组）
我们知道：

temps = “1 2 3”; <=> temps[5] = {“1”, ” “, “2”, ” “, “3”};

所以一开始我就在想：如何跳掉空格，将纯数字储存到数组array[ ]中

int closest = 5526; //温度的范围：-273 ~ 5526
for (int i = 0; i < 3*n; i++) //想到一个数有：符号+数字+空格（**想当然**3*n）
{
int ascii = temps[i];
if (ascii > 47 && ascii < 58) //0 ~ 9 (ascii码 048 ~ 057)
{
if (closest > temps[i]) closest = temps[i];
}
}

上面虽然没有使用数组，但也是这个思路，但我忘了
string型是一个数字一个空间，而不是一个数一个空间……
所以，它最多读取都是一位数的正项数列
虽然也想过拼接在一起，符号、多位数字分别储存，但这样步骤十分麻烦，而且容易出错

后来，仔细的查阅了资料“C++ string转化为int数组”

c++ string转化为int数组
Split a string in C++

//就是这个：
#include <iostream>
#include <sstream>   //字符串流，常用与格式转换
#include <string>

int main()
{
    std::string s("Somewhere down the road");
    std::istringstream iss(s);

    do
    {
        std::string sub;        //可以用int数组，得提前定义
     iss >> sub;             //将值流向sub
       std::cout << "Substring: " << sub << std::endl;
  } while (iss);
}
/*
Output: 
Substring: Somewhere
Substring: down
Substring: the
Substring: road */

上面模板可以改成适合题目的：

int n = 5;                        //the number of temperatures
string temps = "1 -2 -8 4 5";
istringstream out(temps);
int array[n];
for (int i= 0; i < n; i++) {
     out >> array[i];
     cout << array[i] << " ";
}

Output: 1 -2 -8 4 5

通关程序

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <sstream>

using namespace std;

/**
 * Auto-generated code below aims at helping you parse
 * the standard input according to the problem statement.
 **/
int main()
{
    int n, p, q, closest = 5526; // the number of temperatures to analyse
    cin >> n; cin.ignore();
    string temps; 
    // the n temperatures expressed as integers ranging from -273 to 5526
    getline(cin, temps);

    istringstream out (temps);
    int temp[n];

    if ( n == 0) cout << "0" << endl;
    else {
        for (int i = 0; i < n; i++) {
        out >> temp[i];

        if (temp[i] < 0) p = 0 - temp[i];
        else p = temp[i];                 //绝对值

        if (closest < 0) q = -closest;
        else q = closest;                 //绝对值

        if (q > p) closest = temp[i];
        else if (q == p) closest = q;
        }

    // Write an action using cout. DON'T FORGET THE "<< endl"
    cout << closest << endl;
    }
    // To debug: cerr << "Debug messages..." << endl;
    cerr << "Debug messages are " << n << " " << temps << endl;

}

带函数function版
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <sstream>

using namespace std;

int main()
{
    int n;
    cin >> n; cin.ignore();
    string temps; 
    getline(cin, temps);

    int output(int, string);

    cout << output(n,temps) << endl;
}
int output(int num, string temperatures)
{
    istringstream out (temperatures);
    int temp[num], closest = 5526, q, p;

    if ( num == 0) return 0;
    else {
        for (int i = 0; i < num; i++) {

            out >> temp[i];

            if (temp[i] < 0) p = 0 - temp[i];
            else p = temp[i];

            if (closest < 0) q = -closest;
            else q = closest;

            if (q > p) closest = temp[i];
            else if (q == p) closest = q;
        }
        return closest;
    }
}

//成功

//题目全部解决后；后期提交游戏时，出现小问题（07）……好像是因为，如果两个最接近0的数都为负数，但我的程序确实只会输出正数……

试着改了下：

if ( num == 0) return 0;
    else {
        for (int i = 0; i < num; i++) {

            out >> temp[i];

            if (temp[i] < 0) p = 0 - temp[i];
            else p = temp[i];

            if (closest < 0) q = -closest;
            else q = closest;

            if (q > p) closest = temp[i];
            //加上了temp[i]与closest正负的判定
            else if (q == p&& temp[i]*closest < 0) closest = q;
            else if (q == p&& temp[i]*closest > 0) closest = -q;
        }
        return closest;

//这时候，原来问题解决了，但又出现新的问题（03）……
//巧合是，我将修改的第二行去掉，运行成功了
            if (q > p) closest = temp[i];
            //只保留下面这一句
            else if (q == p&& temp[i]*closest < 0) closest = q;