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在 SPOJ 上提交代码会导致运行时错误 (SIGABRT)

[英]Code submission on SPOJ gives runtime error (SIGABRT)

I have done an exercise on SPOJ to practice advanced algorithms.我在 SPOJ 上做了一个练习来练习高级算法。


The Problem Statement is as follows:问题陈述如下:

Harish went to a supermarket to buy exactly 'k' kilograms apples for his 'n' friends.哈里什去超市为他的“n”个朋友买了“k”公斤的苹果。 The supermarket was really weird.超市真的很奇怪。 The pricing of items was very different.物品的定价非常不同。 He went to the Apples section and enquired about the prices.他去苹果区询问价格。 The salesman gave him a card in which he found that the prices of apples were not per kg.售货员给了他一张卡片,他发现苹果的价格不是每公斤。 The apples were packed into covers, each containing 'x' kg of apples, x > 0 and 'x' is an integer.苹果被包装在盖子中,每个包含“x”公斤苹果,x > 0,“x”是 integer。 An 'x' kg packet would be valued at 'y' rupees.一个“x”公斤的小包价值为“y”卢比。 So, the placard contained a table with an entry 'y' denoting the price of an 'x' kg packet.因此,标语牌包含一个表格,其中有一个条目“y”,表示“x”公斤包的价格。 If 'y' is -1 it means that the corresponding packet is not available.如果 'y' 为 -1,则表示相应的数据包不可用。 Now as apples are available only in packets, he decides to buy at most 'n' packets for his 'n' friends ie he will not buy more than n packets of apples.现在由于苹果只能以小包的形式提供,他决定最多为他的“n”个朋友购买“n”包,即他不会购买超过 n 包的苹果。 Harish likes his friends a lot and so he does not want to disappoint his friends. Harish 非常喜欢他的朋友,所以他不想让他的朋友失望。 So now, he will tell you how many friends he has and you have to tell him the minimum amount of money he has to spend for his friends.所以现在,他会告诉你他有多少朋友,你必须告诉他他必须为朋友花的最低金额。


This is the code I used to solve the problem:这是我用来解决问题的代码:

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

using std::cout;
using std::cin;
using std::vector;
using std::endl;

int MinValueOf(int a, int b)
{
    return (a < b) ? a : b;
}
int BuyingApple(vector<int> PriceaTag, int Friends, int KilogramsToBuy)
{
    vector<vector<int>> Table(Friends + 1, vector<int>(KilogramsToBuy + 1, 0));
    for(int i = 1; i <= Friends; i++)
    {
        for(int j = 0; j <= i; j++)
        {
            Table[i][j] = INT32_MAX;
            if(j == 0)
                Table[i][0] = 0;
            else if(PriceaTag[j] > 0)
                Table[i][j] = MinValueOf(Table[i][j], Table[i - 1][i - j] +  PriceaTag[j]);
        }
    }
    return (Table[Friends][KilogramsToBuy] == 0) ? -1 : Table[Friends][KilogramsToBuy];
}
int main()
{
    vector<int> Price;
    int Friends, Kilogram, t;
    cin >> t;
    for(int i = 0; i < t; i++)
    {
        cin >> Friends >> Kilogram;
        vector<int> Price(Kilogram + 1, 0);
        for(int i = 1; i <= Kilogram; i++)
        {
            cin >> Price[i];
        }
        cout << BuyingApple(Price, Friends, Price.size() - 1) << endl;
    }
    return 0;
}

I/O of the code is as follows:代码的I/O如下:

The first line of input will contain the number of test cases, C.第一行输入将包含测试用例的数量,C。 Each test case will contain two lines.每个测试用例将包含两行。 The first line containing N and K, the number of friends he has and the amount of Apples in kilograms which he should buy.第一行包含 N 和 K,他的朋友数量和他应该购买的苹果数量(公斤)。 The second line contains K space separated integers in which the ith integer specifies the price of a 'i' kg apple packet.第二行包含 K 个空格分隔的整数,其中第 i 个 integer 指定“i”公斤苹果包的价格。 A value of -1 denotes that the corresponding packet is unavailable. -1 值表示对应的数据包不可用。

The output for each test case should be a single line containing the minimum amount of money he has to spend for his friends.每个测试用例的 output 应该是一行,其中包含他必须为他的朋友花费的最低金额。 Print -1 if it is not possible for him to satisfy his friends.如果他不可能满足他的朋友,则打印 -1。


Constraints:约束:

0 < N <= 100 0 < N <= 100
0 < K <= 100 0 < K <= 100
0 < price <= 1000 0 < 价格 <= 1000


But as I submitted my code, I received a message SIGABRT runtime error although my code ran smoothly in both Windows compiler (G++ 14) and Linux Compiler (G++ Clang 9) .但是当我提交我的代码时,我收到一条消息SIGABRT runtime error ,尽管我的代码在Windows compiler (G++ 14)Linux Compiler (G++ Clang 9)中运行顺利。 I have tried to debug but I failed.我试图调试,但我失败了。 What might be wrong?可能有什么问题?

Since this is a SPOJ question, and you're not given the test data, what you should do is to randomize the tests until you get a failure.由于这是一个 SPOJ 问题,并且您没有获得测试数据,因此您应该做的是随机化测试,直到您失败。 That way, you may be able to get a sample case that fails.这样,您可能会得到一个失败的示例案例。 This is called fuzzing , and is a technique that can be used in your question.这称为fuzzing ,是一种可以在您的问题中使用的技术。

The following will work for the cases that cause segmentation faults, and in some cases, to verify if a given output matches the expected output.以下将适用于导致分段错误的情况,并且在某些情况下,验证给定的 output 是否与预期的 output 匹配。 In other words, instead of trying to figure out the test data, let the computer generate the tests for you.换句话说,与其试图找出测试数据,不如让计算机为您生成测试。

The way you do this is to look at the constraints that the question gives you, and generate random data that fits the constraints.你这样做的方法是查看问题给你的约束,并生成符合约束的随机数据。 Since they are all integers, you can do this by using the <random> header, and using a uniform_int_distribution .由于它们都是整数,因此您可以使用<random> header 并使用uniform_int_distribution来做到这一点。

Here is a sample of fuzzing the data using the following constraints for N , K , and the data for prices:这是使用以下NK约束和价格数据对数据进行模糊测试的示例:

Constraints:

0 < N <= 100
0 < K <= 100
0 < price <= 1000

OK, so given this information, we can take your exact code, remove the cin statements, and replace everything with randomized data that fit the constraints.好的,鉴于此信息,我们可以获取您的确切代码,删除cin语句,并将所有内容替换为符合约束的随机数据。 In addition, we can test for an out-of-bounds access if we use at() to access the vectors in the function that causes the issue.此外,如果我们使用at()访问导致问题的 function 中的向量,我们可以测试是否存在越界访问。

Given all of this information we can start with changing main to produce random data that fit the constraints of the question:鉴于所有这些信息,我们可以从更改main开始以生成符合问题约束的随机数据:

#include <random>
#include <algorithm>
//...
int main()
{
    // random number generator
    std::random_device rd;
    std::mt19937 gen(rd());

    // Prices will be distributed from -1 to 1000
    std::uniform_int_distribution<> distrib(-1, 1000);

    // N and K are distributed between 1 and 100  
    std::uniform_int_distribution<> distribNK(1, 100);

    // This one will be used if we need to replace 0 in the Price vector with 
    // a good value 
    std::uniform_int_distribution<> distribPos(1, 1000);

    // our variables
    int Friends;
    int Kilogram;
    vector<int> Price;

    // we will keep going until we see an out-of-range failure
    while (true)
    {
        try
        {
            // generate random N and K values
            Friends = distribNK(gen);
            Kilogram = distribNK(gen);

            // Set up the Price vector
            Price = std::vector<int>(Kilogram + 1, 0);

            // Generate all the prices
            std::generate(Price.begin() + 1, Price.end(), [&]() { return distrib(gen); });

            // Make sure we get rid of any 0 prices and replace them with a random value
            std::transform(Price.begin() + 1, Price.end(), Price.begin() + 1, [&](int n)
                { if (n == 0) return distribPos(gen);  return n; });

            // Now test the function
            std::cout << BuyingApple(Price, Friends, Price.size() - 1) << std::endl;
        }

        catch (std::out_of_range& rError)
        {
            std::cout << rError.what() << "\n";
            std::cout << "The following tests cause an issue:\n\n";
            // The following tests cause an issue with an out-of-range.  See the data
            std::cout << "Friends = " << Friends << "\nK = " << Kilogram << "\nPrice data:\n";
            int i = 0;
            for (auto p : Price)
            {
                std::cout << "[" << i << "]: " << p << "\n";
                ++i;
            }
            return 0;
        }
    }
}

Given all of this, we can change the BuyingApple function by replacing [ ] with at() :鉴于所有这些,我们可以通过将[ ]替换为at()来更改BuyingApple function:

int BuyingApple(vector<int> PriceaTag, int Friends, int KilogramsToBuy)
{
    vector<vector<int>> Table(Friends + 1, vector<int>(KilogramsToBuy + 1, 0));
    for (int i = 1; i <= Friends; i++)
    {
        for (int j = 0; j <= i; j++)
        {
            Table.at(i).at(j) = INT32_MAX;
            if (j == 0)
                Table[i][0] = 0;
            else if (PriceaTag[j] > 0)
                Table[i][j] = MinValueOf(Table[i][j], Table.at(i - 1).at(i - j) + PriceaTag.at(j));
        }
    }
    return (Table[Friends][KilogramsToBuy] == 0) ? -1 : Table[Friends][KilogramsToBuy];
}

Now we have an automatic case generator, and will catch and display any cases that could cause an issue with the vectors.现在我们有一个自动案例生成器,它将捕获并显示任何可能导致向量出现问题的案例。 Note that we keep looping forever until we get a test case that "crashes".请注意,我们会一直循环下去,直到我们得到一个“崩溃”的测试用例。 We then output the crashed case and can now use those values to debug the issue.然后我们 output 崩溃案例,现在可以使用这些值来调试问题。

We used std::generate and std::transform as an illustration of how to populate a vector (or any sequence container your test uses), and how to specialize the test (like making sure that Price has no 0 values).我们使用std::generatestd::transform来说明如何填充向量(或您的测试使用的任何序列容器),以及如何专门化测试(例如确保Price没有0值)。 Another SPOJ question may need other specializations, but hopefully you get the basic idea.另一个 SPOJ 问题可能需要其他专业知识,但希望您了解基本概念。

Here is a Live Example .这是一个活生生的例子

We see that a test case caused an out-of-range exception to be thrown.我们看到一个测试用例导致抛出一个out-of-range异常。 The main function has a try/catch to process this error, and we can see the data that caused the issue. main的 function 有一个try/catch来处理这个错误,我们可以看到导致问题的数据。


So it seems that if we have more friends than apples, the issue occurs where we go out-of-bounds.所以看起来如果我们的朋友比苹果多,问题就会出现在我们 go 越界的地方。 I will not attempt to fix the issue, but you now have a test case where the input fails.我不会尝试解决这个问题,但您现在有一个输入失败的测试用例。

In general, you can use this technique with many, if not most of the "online judge" sites if the site doesn't show you failing test cases.一般来说,如果网站没有向您显示失败的测试用例,您可以将这种技术用于许多(如果不是大多数)“在线法官”网站。

Edit: Updated the lambda in the std::transform to only replace 0 in the Price vector.编辑:更新了std::transform中的 lambda 以仅替换Price向量中的0


Edit: Here is a random string fuzzer that can produce fuzzed string data.编辑:这是一个随机字符串模糊器,可以生成模糊字符串数据。

You can control the number of strings, the minimum and maximum size of each string, and the alphabet of characters that will be used when generating each string.您可以控制字符串的数量、每个字符串的最小和最大大小以及生成每个字符串时将使用的字符字母表。

#include <random>
#include <string>
#include <vector>
#include <iostream>

struct StringFuzzer
{ 
    unsigned int maxStrings;  // number of strings to produce
    unsigned int minSize;     // minimum size of a string
    unsigned int maxSize;     // maximum size of the string
    bool fixedSize;           // Use fixed size of strings or random
    std::string alphabet;     // string alphabet/dictionary to use
    
public:
    StringFuzzer() : maxStrings(10), minSize(0), maxSize(10), fixedSize(true), alphabet("abcdefghijklmnopqrstuvwxyz")
    {}
    StringFuzzer& setMaxStrings(unsigned int val) { maxStrings = val; return *this; };
    StringFuzzer& setMinSize(unsigned int val) { minSize = val; return *this; };
    StringFuzzer& setMaxSize(unsigned int val) { maxSize = val; return *this; };
    StringFuzzer& setAlphabet(const std::string& val) { alphabet = val; return *this; };
    StringFuzzer& setFixedSize(bool fixedsize) { fixedSize = fixedsize; return *this; }

    std::vector<std::string> getFuzzData() const
    {
        // random number generator
        std::random_device rd;
        std::mt19937 gen(rd());

        // Number of strings produced will be between 1 and maxStrings
        std::uniform_int_distribution<> distribStrings(1, maxStrings);

        // string size will be distributed between min and max sizes
        std::uniform_int_distribution<> distribMinMax(minSize, maxSize);

        // Picks letter from dictionary
        std::uniform_int_distribution<> distribPos(0, alphabet.size() - 1);

        std::vector<std::string> ret;

        // generate random number of strings
        unsigned int numStrings = maxStrings;
        if ( !fixedSize)
           numStrings = distribStrings(gen);
           
        ret.resize(numStrings);

        for (unsigned int i = 0; i < numStrings; ++i)
        {
            std::string& backStr = ret[i];
            // Generate size of string
            unsigned strSize = distribMinMax(gen);
            for (unsigned j = 0; j < strSize; ++j)
            {
                // generate each character and append to string
                unsigned pickVal = distribPos(gen);
                backStr += alphabet[pickVal];
            }
        }
        return ret;
    }
};

int main()
{
    StringFuzzer info;
    auto v = info.getFuzzData();  // produces a vector of strings, ready to be used in tests
    info.setAlphabet("ABCDEFG").setMinSize(1);  // alphabet consists only of these characters, and we will not have any empty strings
    v = info.getFuzzData();  // now should be a vector of random strings with "ABCDEFG" characters
    for (auto s : v)
       std::cout << s << "\n";
}

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