Question

我有一份家庭作业，我觉得我已经接近正确。作业如下：

银行互相贷款。在经济困难时期，如果银行破产，它可能无法偿还贷款。银行的总资产是其当前余额加上对其他银行的贷款。图8.1（附图）是一个图表，显示了五个银行。银行目前的余额分别为：25,125,175,75和1.81亿美元。从节点1到节点2的有向边指示银行1贷款40到银行2。

diagram

如果银行总资产低于一定限额，银行将被视为不安全。如果银行不安全，其借入的资金不能退还给贷方，贷方也不能将贷款计入其总资产中。因此，贷款人也可能不安全。

编写程序以查找所有不安全的银行。您的程序按如下方式读取输入。它首先读取两个整数n和limit，其中n表示银行数量，limit是保持银行安全的最低资产。然后，它会读取n行，这些行描述id为0到n-1的n个银行的信息。该行中的第一个数字是银行的余额。第二个数字表示从银行借来的钱数，其余的是两个数字对。每对描述一个借款人。第一个数字是银行ID，第二个数字是借入的数量。假设最大存储体数为100.例如，五个存储体的输入如下（限制为201）

5 201

25 2 1 100.5 4 320.5

125 2 2 40 3 85

175 2 0 125 3 75

75 1 0 125

181 1 2 125

银行3的总资产为75加125，不到201，因此银行不安全。在银行3不安全之后，银行1的总资产变为125 + 40，现在也不安全。输出应该是“不安全的银行是3 1”

这是我目前解决问题的方法。我无法弄清楚如何找到所有不安全的银行。只是第一个。我把它设置为静态输入进行测试。如果我能让它正常工作，我就可以使用工作用户输入代码了。

#include <iostream>

using namespace std;

int main()
{
    const int SIZE = 100;
    double balance[SIZE];
    double loan[SIZE][SIZE];
    int nobanks;
    int limit;
    int i = 0;
    int j = 0;
    int k = 0;
    int noborrowers;
    double assets[SIZE];
    bool isSafe[SIZE];
    bool newunsafefound = true;

    cout << "Enter number of banks and the limit:" << endl;

    // Set all of the data
    nobanks = 5;
    limit = 201;
    balance[0] = 25.0;
    balance[1] = 125.0;
    balance[2] = 175.0;
    balance[3] = 75.0;
    balance[4] = 181.0;
    loan[0][1] = 100.5;
    loan[0][4] = 320.5;
    loan[1][2] = 40.0;
    loan[1][3] = 85.0;
    loan[2][0] = 125.0;
    loan[2][3] = 75.0;
    loan[3][0] = 125.0;
    loan[4][2] = 125.0;

    // Set array to all true values
    for(i = 0; i < nobanks; i++)
    {
        isSafe[i] = true ;
    }

    cout << "Unsafe banks are: ";

    i=0;

    while(isSafe[i] == true)
    {
        newunsafefound=false;

        i=0;

        do
        {
            assets[i] = balance[i]; //Set assets to balance

            for (j = 0; j < nobanks; j++) // Check if a bank has loans and add them to assets
            {
                if (loan[i][j] >= 0)
                    assets[i] += loan[i][j];
            }

            if (assets[i] < limit) // Check to see if current bank meets limit
            {
                isSafe[i] = false; // Set bank to not safe if limit not met

                newunsafefound = true;

                cout << i << " " ; //Display the bank that is unsafe and a space for the next bank

                k=0;

                for (k = 0; k < nobanks; k++)
                {
                    loan[i][k] = 0; //Set banks loans to 0 if unsafe.
                    k++;
                }
            }
            i++;
        } while(i < nobanks);
    }
    return (0);
}

我做错了什么？

Answer 1

for (k = 0; k < nobanks; k++)
{
    loan[i][k] = 0; //Set banks loans to 0 if unsafe.
    k++;
}

k的额外增量看起来非常可疑;）

Answer 2

您必须显式初始化loan数组，因此您不使用的元素没有任意值：

double loan[SIZE][SIZE] = {{0}};

同样loan[i][k] = 0;表示您将银行i 给予的贷款归零到银行k，但您想要的是零钱银行i 借用到银行k。

还有一个问题：

在外部循环的退出条件下（只有在最后一个银行(nobank-1)不安全时才会退出），
处理newunsafefound时没有做变量名称所暗示的内容。

Answer 3

// Astrid Giraldo

public static void main (String[] args) {

    Scanner input = new Scanner(System.in);

    int n;               // number of banks
    double limit;        // Minimum total assets for keeping a bank safe.

    System.out.println("Enter number of banks:  ");
    n = input.nextInt();

    System.out.println("Enter minimum total assets to be a safe bank:  ");
    limit = input.nextDouble();

    double[][] borrowers = new double[n][n];        
    double[] balance = new double[n];
    int numBorrowers;

    for (int i = 0; i < borrowers.length; i++) {
        System.out.println("Enter the bank's balance");
        balance[i] = input.nextDouble();
        System.out.println("Enter number of borrowers from this bank");
        numBorrowers = input.nextInt();

        for (int j = 0; j < numBorrowers; j++) {
            System.out.println("Enter borrower bank id and the amount borrowed:  ");
            borrowers[i][input.nextInt()] = input.nextDouble();

        }
    }
    markUnsafeBanks(borrowers,balance, limit);
    displayUnsafeBanks(borrowers, balance, limit);
}



public static double analizeUnsafeBanks(double[][] borrowers, double[] balance,  int bankId) {
    double sum = balance[bankId];

    for (int i = 0; i < borrowers.length; i++) {
        sum += borrowers[bankId][i];
    }

    return sum;
} 

public static void setLoanToZero(double[][] borrowers, double[] balance, int unsafeBankId, double limit) {

    for (int i = 0; i < borrowers.length; i++) {
        if (borrowers[i][unsafeBankId] > 0) {
            borrowers[i][unsafeBankId] = 0;
            if ( unsafeBankId > i && analizeUnsafeBanks(borrowers, balance, i) < limit) {
                setLoanToZero(borrowers, balance, i, limit);
            }
        }
    }
}

public static void markUnsafeBanks(double[][] borrowers, double[] balance, double limit) {
    for (int i = 0; i < borrowers.length; i++) {
        if (analizeUnsafeBanks(borrowers, balance, i) < limit) {
            setLoanToZero(borrowers, balance, i, limit);
        }
    }
}

public static void displayUnsafeBanks(double[][] borrowers, double[] balance, double limit){
    for (int i = 0; i < borrowers.length; i++) {
        double assets = analizeUnsafeBanks(borrowers,balance,i);
        if ( assets < limit) {
            System.out.println("Bank " + i + " is unsafe.  It assets are " + assets);
        }   
    }   
}

循环问题，过早停止

3 个答案: