[英]Searching for the upper bound in an array
我正在尝试使用变量sum
的值作为上限来获取数组size
最近上限的索引 (除非找到等效值),然后在数组value
的相同索引处找到该value
。
例如:如果在值sum
是270,我的节目应该找到在位于索引6的值280 size
和输出在对应的值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;
}
将您的代码更改为此 -
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";
}
}
还有其他方法可以解决这个问题。 但正如你所说,你是初学者。 我给出了简单的强力解决方案。 :)
要获取上限的索引,只需像这样使用std::upper_bound
(要求范围至少部分排序):
// 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);
然后使用index
例如:
std::cout << value[index] << std::endl;
最简单的方法可以分为两行:
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;
请注意, size
必须按总和进行分区。 您的示例中的排序数组符合此条件。
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