Question

我有以下C ++程序：

tail

我使用它作为输入（在文件“input.txt”中）：

#include "stdafx.h"
#include <iostream>
#include <fstream>
#include <ostream>

using namespace std;
ifstream input("input.txt");
long long N, C, D;
long long a[100000][4];
long long b[100000][4];
int case_number = 0;
long long chosen_machine;

long long max(long long v1, long long v2)
{
    // returns the maximume value between 2 values
    if (v1 > v2)
        return v1;
    else
        return v2;
}

long long f_recursive(long long cash, long long i, long long i_chosen, int e)
{
    if (i == N - 1)
    {
        if (e == 0) //we dont buy the current machine
        {
            return (cash + (D-b[i][0]+1)*b[i_chosen][3] );
        }
        else //e==1 and we will buy the current machine
        {
            if (cash < b[i][1]) //we cant buy it
            {
                if (i_chosen == -1)
                    return cash;
                else
                    return (cash + (D - b[i][0] + 1)*b[i_chosen][3]);
            }
            cash = cash - b[i][1] + b[i][2];
            return (cash + (D - b[i][0])*b[i][3]);
        }
    }

    if (e == 1) //we are going to buy the machine
    {
        if (cash < b[i][1])
        {
            //we cant buy it so we keep the current cash and go to next machine
            return max(f_recursive(cash, i + 1, i_chosen, 1), f_recursive(cash, i + 1, i_chosen, 0));
        }
        cash = cash - b[i][1] + b[i][2] + (b[i + 1][0] - b[i][0] - 1)*b[i][3];
        return max(f_recursive(cash, i + 1, i, 1), f_recursive(cash, i + 1, i, 0));
    }
    else // e==0 and we wont buy the machnie
    {
        if (i_chosen == -1) // if we did not buy any machine before (we still own no machine)
        {
            return max(f_recursive(cash, i + 1, i_chosen, 1), f_recursive(cash, i + 1, i_chosen, 0));
        }
        cash = cash + (b[i + 1][0] - b[i][0])*b[i_chosen][3];
        return max(f_recursive(cash, i + 1, i_chosen, 1), f_recursive(cash, i + 1, i_chosen, 0));
    }
}

int read_case()
{
    //reads each case and corresponding N C D and also the machines details to array a
    string line;
    input >> N >> C >> D;
    for (long long i = 0; i < N; i++)
    {
        input >> a[i][0] >> a[i][1] >> a[i][2] >> a[i][3];
    }
    return 1;
}

int sort_a()
{
    long long checked[100000];
    for (long long i = 0; i < N; i++)
    {
        long long max = 999999999999999999;
        long long j_chosen;
        for (long long j = 0; j < N; j++)
        {
            if (a[j][0] <= max && checked[j]!=1)
            {
                max = a[j][0];
                j_chosen = j;
            }
        }
        b[i][0] = a[j_chosen][0];
        b[i][1] = a[j_chosen][1];
        b[i][2] = a[j_chosen][2];
        b[i][3] = a[j_chosen][3];
        checked[j_chosen] = 1;
    }
    return 1;
}

int main()
{
    int exit_condition = 0;
    while (!exit_condition)
    {
        read_case();
        if (N == 0 && C == 0 && D == 0)
        {
            exit_condition = 1;
            continue;
        }

        if (N==0) // there is no machine, so simply output the dollar we have
            cout << "Case " << ++case_number << ": " << C << "\n";
        else
        {
            sort_a();

            cout << "Case " << ++case_number << ": " << max(f_recursive(C, 0, -1, 0), f_recursive(C, 0, -1, 1)) << "\n";
        }
    }       

    input.close();
    cout << "Finished, please press Enter...";
    getchar();
    return 0;
}

在Microsoft Visual Studio（C ++）中，代码生成此输出（它在Debug或Release上执行此操作）：

6 10 20
6 12 1 3
1 9 1 2
3 2 1 2
8 20 5 4
4 11 7 4
2 10 9 1
4 979142410 1000000000
719554784 535235252 43212142 2
679922882 929213070 72282699 3
947550084 114776168 96759913 1
811283493 865444308 248312924 4
10 936980155 1000000000
762761567 683459202 109944378 6
686089795 734650986 45620682 4
678254540 925086416 24622261 5
807652386 596279745 140024880 9
791494729 585405092 50583956 8
655401693 691328196 82491162 3
799450676 654579438 65612488 10
580278950 624801431 68664589 2
834332323 692933073 524827737 7
101802074 760182347 3637050 1
1 10 100
50 11 9 100
1 10 100
50 10 9 100
1 10 100
1 10 9 100
1 10 100
100 10 9 100
1 10 100
95 10 1 1
1 10 100
95 10 1 2
1 1000000000 1000000000
1 10 9 1000000000
2 10 100
80 67 60 7
20 8 5 1
2 10 100
80 66 60 7
20 8 5 1
2 10 100
80 66 60 1
20 8 5 7
2 10 30
20 8 5 8
20 2 1 7
0 0 0

在Linux（gcc g ++）上，它产生这个输出：

Case 1: 44
Case 2: 1116877054
Case 3: 2353506445
Case 4: 10
Case 5: 5009
Case 6: 9909
Case 7: 10
Case 8: 10
Case 9: 11
Case 10: 999999999999999999
Case 11: 87
Case 12: 200
Case 13: 567
Case 14: 87
Finished, please press Enter..

作为参考，我试图解决这个问题： http://docdro.id/H6fd5bs

有人知道或不知道输入和输出在不同的编译器上会有什么不同吗？

Answer 1

在visual studio中创建变量时，默认情况下其值为零，但在Linux g ++中，变量将是随机的。因此，分配变量将解决这个问题。这是我遇到的问题，但是，基于代码会有其他问题。

为什么visual studio和gcc为c ++代码生成不同的输出

1 个答案: