搜索数组的上限

Question

I'm trying to get the index of the nearest upper bound (unless equivalent value found) in the array size using the value in variable sum as upper bound, and then find the value at the same index in the array value .

For example: if the value in sum is 270, my program should find the value 280 located at index 6 in size and output the value at corresponding value[6] .

#include <iostream>
#include <cmath>
#include <cstring>

using namespace std;

int main()
{
    double x = 0;
    double y = 0;
    double sum = 0;
    double size[27] = {24, 28, 32, 38, 48, 240, 280, 320, 360, 380,
                       420, 480, 560, 600, 640, 700, 720, 800, 840,
                       960, 980, 1120, 1200, 1280, 1440, 1680, 1920};

    double value[27] = {.0022, .0026, .0029, .0035, .0044, .0219,
                        .0256, .0292, .0328, .0384, .0438, .0513,
                        .0547, .0584, .0641,.0656, .073, .0766,
                        .0875, .0877, .0897, .1023, .1094, .1169,
                        .1313, .1531, .175};

    cout << "Enter width: " << endl;
    cin >> x;
    cout << "Enter height: " << endl;
    cin >> y;

    x = ceil(x) + 3;
    y = ceil(y) + 3;

    sum = x * y;
}

Answer 1

Change your code to this -

    double x = 0;
    double y = 0;
    double sum = 0;
    int size[27] = {24, 28, 32, 38, 48, 240, 280, 320, 360, 380,
    420, 480, 560, 600, 640, 700, 720, 800, 840, 960, 980, 1120, 1200, 1280, 1440, 1680, 1920};
    double value[27] = {.0022, .0026, .0029, .0035, .0044, .0219,
    .0256, .0292, .0328, .0384, .0438, .0513, .0547, .0584, .0641,.0656, .073, .0766, .0875, .0877, .0897, .1023, .1094, .1169, .1313, .1531, .175};

    cout << "Enter width: " << endl;
    cin >> x;
    cout << "Enter height: " << endl;
    cin >> y;

    x = ceil(x) + 3;
    y = ceil(y) + 3;

    sum = x * y;

    for (int i=0;i<27;i++)
    {
        if (size[i]>=sum)
        {
          cout<<value[i]<<endl; 
          break;
        }
        else if(i==26)
        {
            cout<<"No upper Bound find\n";
        }
    }

There are other ways to solve this. But as you said you are a beginner. I have given the simple bruteforce solution. :)

Answer 2

To get the index of the upper bound simply use std::upper_bound like this (requires that the range is at least partially sorted):

// Get iterator to upper bound.
auto it = std::upper_bound(std::begin(size), std::end(size), sum);

// Get index by iterator subtraction.
std::size_t index = it - std::begin(size);

Then use index eg as:

std::cout << value[index] << std::endl;

Answer 3

The simplest way can be done in 2 lines:

auto size_ub = std::upper_bound(std::begin(size), std::end(size), sum);
int idx = std::distance(std::begin(size), size_ub);

cout << value[idx] << endl;

Please note that size must be partitioned with respect to sum. A sorted array as in your example meets this criteria.