我的函数中的百分比计算不起作用

Question

所以我要做的是取一个数字向量，找出每个元素出现的时间，然后计算其在向量中出现次数的百分比当量。

它在只有一个数的几个元素的向量上非常有效，但在多次出现的不同元素(确切地说是2(的向量上则完全无效。

这是一个详细的输出：

count = 6
pushed back 21
total = 6
percent = 6 / 6* 100
pushed back 100
Number:
21 
Percent Chance
100 
count = 5
pushed back 42
count = 5
pushed back 21
total = 5
total = 10
percent = 5 / 10* 100
pushed back 0
percent = 5 / 10* 100
pushed back 0
Number:
42 
Percent Chance
0 // SHOULD BE 50
Number:
21 
Percent Chance
0 // SHOULD BE 50

这是代码：(输出来自该函数，使用不同的x矢量运行两次(

std::vector <std::string>  findpercentages(std::vector <std::string> x)
{
int count{};                          //for amount of times element occurs
std::vector <std::string> tempvec{};  //to hold the element/s
std::vector <int> counts{};           //to hold count for each element in tempvec
std::vector <std::string> finalvec{}; //elements and counts of elements
for (size_t i = 0; i < x.size(); i++)
{
//look for x[element] in tempvec, if not in tempvec, count occurances in x and add element to 
// tempvec, add count to counts 
std::vector <std::string>::iterator point { std::find(tempvec.begin(), tempvec.end(), x[i]) };
if (point == tempvec.end())
{
count = std::count(x.begin(), x.end(), x[i]);
std::cout << "count = " << count << 'n';
counts.push_back(count);
std::cout << "pushed back " << x[i] << 'n';
tempvec.push_back(x[i]);
}
}
int total{};
for (size_t n = 0; n < counts.size(); n++)
{
total += counts[n]; //total for percentage calculation
std::cout << "total = " << total << 'n';
}
for (size_t y = 0; y < tempvec.size(); y++)
{
finalvec.push_back(tempvec[y]);
//percent calculation. used unsigned_int64 because got arithmatic overflow warning if i just use 
// int
unsigned __int64 percentage = static_cast <unsigned __int64> (round((counts[y] / total) * 100));
std::cout << "percent = " << counts[y] << " / " << total << "* 100" << 'n';
finalvec.push_back(std::to_string(percentage));
std::cout << "pushed back " << percentage << 'n';
}
return finalvec;
}

Answer 1

unsigned __int64 percentage = static_cast <unsigned __int64> (round((counts[y] / total) * 100));

应该是类似的东西

double percentage = round((100.0 * counts[y]) / total);

你的问题是，当你用一个整数除以另一个整数时，你总是得到一个整数。所以(counts[y] / total)总是0或1。

因此，最简单的解决方案是在计算中引入double值100.0。这样可以确保得到浮点除法而不是整数除法。