如何在多维数组中搜索单个行，然后对某些行中的每个单个元素执行计算

我目前正在使用桌子来查找与船只所持有的燃料总量相关的单个储罐中的相应燃料水平。该船目前有8个储罐，并将燃料分配给该船携带的总燃料的量。当给出总燃料量时，我需要在8个储罐中以总燃料量查找或填充燃料的位置。

我已经尝试搜索2D阵列以在给定时间查找相应的燃油箱，但是使用图表(2D数组(，我意识到我需要为左列使用索引(这是总燃料总数((，然后找到下边界和上限行。之后，我需要找到每个单个元素(燃油箱(的平均值。我能够发表数学的语句和硬核，但是我知道有一种递归方法可以搜索左列，找到给定燃料量的上和下键行，并在每个燃油箱中找到量。因此，如果我给出了IE的目标量。194568，我需要大约找到8个储罐中的燃料在哪里。

int main()
const int NUM_ROWS = 22;
    const int NUM_COLUMNS = 9;
    int arr[NUM_ROWS][NUM_COLUMNS] = {
    {80000,4500,7500,11700,11700,0,0,0,9200},
    {90000,4500,7500,13800,13800,0,0,0,10800 },
    {100000,4500,7500,15900,15900,0,0,0,12400},
    {110000,4500,7500,18200,18200,0,0,0,13200},
    {120000,4500,7500,20400,20400,0,0,0,14400},
    {130000,4500,7500,22000,23000,0,0,0,16000},
    {140000,4500,7500,22000,26200,0,0,0,19600},
    {150000,4500,7500,22000,29400,0,0,0,23200},
    {160000,4500,7500,22000,32700,0,0,0,26600},
    {170000,4500,7500,23200,35400,0,0,0,28800},
    {180000,4500,7500,25300,37500,0,0,0,30400},
    {190000,4500,7500,27400,39600,0,0,0,32000},
    {200000,4500,7500,29500,41700,0,0,0,33600},
    {210000,4500,7500,31800,44300,0,0,0,33800},
    {220000,4500,7500,31800,44300,0,0,10000,33800},
    {230000,4500,7500,31800,44300,0,3000,15000,35800},
    {240000,4500,7500,31800,44300,0,8000,15000,40800},
    {250000,4500,7500,31800,44300,0,8000,25000,40800},
    {260000,4500,7500,31800,44300,0,8000,35000,40800},
    {270000,4500,7500,31800,44300,0,8000,45000,40800},
    {280000,4500,7500,31800,44300,0,14000,45000,44800},
    {290000,4500,7500,31800,44300,0,18800,46000,49000}, };
    int key;
    cout << "Enter Fuel Amount:";
    cin >>  key;
        if (key >= 80000 && key < 90000)
            {

            cout << "L/R_Ext = " << (arr[0][1] + arr[1][1]) / 2 << endl;
                cout << "L/R_ Outbrd = " << (arr[0][2] + arr[1][2]) / 2 << endl;
                cout << "L1 R4 = " << (arr[0][3] + arr[1][3]) / 2 << endl;
                cout << "L2 R3 = " << (arr[0][4] + arr[1][4]) / 2 << endl;
                cout << "Forward = " << (arr[0][5] + arr[1][5]) / 2 << endl;
                cout << "Center = " << (arr[0][6] + arr[1][6]) / 2 << endl;
                cout << "Mid = " << (arr[0][7] + arr[1][7]) / 2 << endl;
                cout << "Aft = " << (arr[0][8] + arr[1][8]) / 2 << endl;
            }
            else if (key >= 90000 && key < 100000)
            {
                cout << "L/R_Ext = " << (arr[1][1] + arr[2][1]) / 2 << endl;
                cout << "L/R_ Outbrd = " << (arr[1][2] + arr[2][2]) / 2 << endl;
                cout << "L1 R4 = " << (arr[1][3] + arr[2][3]) / 2 << endl;
                cout << "L2 R3 = " << (arr[1][4] + arr[2][4]) / 2 << endl;
                cout << "Forward = " << (arr[1][5] + arr[2][5]) / 2 << endl;
                cout << "Center = " << (arr[1][6] + arr[2][6]) / 2 << endl;
                cout << "Mid = " << (arr[1][7] + arr[2][7]) / 2 << endl;
                cout << "Aft = " << (arr[1][8] + arr[2][8]) / 2 << endl;
            }
            else if (key >= 100000 && key < 110000)
            {
                cout << "L/R_Ext = " << (arr[2][1] + arr[3][1]) / 2 << endl;
                cout << "L/R_ Outbrd = " << (arr[2][2] + arr[3][2]) / 2 << endl;
                cout << "L1 R4 = " << (arr[2][3] + arr[3][3]) / 2 << endl;
                cout << "L2 R3 = " << (arr[2][4] + arr[3][4]) / 2 << endl;
                cout << "Forward = " << (arr[2][5] + arr[3][5]) / 2 << endl;
                cout << "Center = " << (arr[2][6] + arr[3][6]) / 2 << endl;
                cout << "Mid = " << (arr[2][7] + arr[3][7]) / 2 << endl;
                cout << "Aft = " << (arr[2][8] + arr[3][8]) / 2 << endl;
            }
    cout << "First row:n";
    for (col = 0; col < NUM_COLUMNS; col++) {
        cout << arr[0][col] << 't';
    }
    cout << "Second row:n";
    for (col = 0; col < NUM_COLUMNS; col++) {
        cout << arr[1][col] << 't';
    }
    cout << "nnSecond column:n";
    for (row = 0; row < NUM_ROWS; row++) {
        cout << arr[row][1] << 'n';
    }
    cout << "nnThird column:n";
    for (row = 0; row < NUM_ROWS; row++) {
        cout << arr[row][2] << 'n';
    }
        cout << "nnForth column:n";
        for (row = 0; row < NUM_ROWS; row++) {
            cout << arr[row][3] << 'n';
        }
/*etc....

我期望有125648的目标时结果会搜索数组并将每个水箱中的燃料返回

int LR_Ext;
int LR_Outbrd;
int L1R4;
int L2R3;
int Forward;
int Center;
int Mid;
int Aft;

也许将它们分配给变量。