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Display a linear combination: Difference between revisions

m (→‎{{header|Factor}}: use when-empty instead of several other words)
Line 131:
Vector for [-1] -> - e1
</pre>
 
=={{header|C++}}==
{{trans|D}}
<lang cpp>#include <iomanip>
#include <iostream>
#include <sstream>
#include <vector>
 
template<typename T>
std::ostream& operator<<(std::ostream& os, const std::vector<T>& v) {
auto it = v.cbegin();
auto end = v.cend();
 
os << '[';
if (it != end) {
os << *it;
it = std::next(it);
}
while (it != end) {
os << ", " << *it;
it = std::next(it);
}
return os << ']';
}
 
std::ostream& operator<<(std::ostream& os, const std::string& s) {
return os << s.c_str();
}
 
std::string linearCombo(const std::vector<int>& c) {
std::stringstream ss;
for (size_t i = 0; i < c.size(); i++) {
int n = c[i];
if (n < 0) {
if (ss.tellp() == 0) {
ss << '-';
} else {
ss << " - ";
}
} else if (n > 0) {
if (ss.tellp() != 0) {
ss << " + ";
}
} else {
continue;
}
 
int av = abs(n);
if (av != 1) {
ss << av << '*';
}
ss << "e(" << i + 1 << ')';
}
if (ss.tellp() == 0) {
return "0";
}
return ss.str();
}
 
int main() {
using namespace std;
 
vector<vector<int>> combos{
{1, 2, 3},
{0, 1, 2, 3},
{1, 0, 3, 4},
{1, 2, 0},
{0, 0, 0},
{0},
{1, 1, 1},
{-1, -1, -1},
{-1, -2, 0, -3},
{-1},
};
 
for (auto& c : combos) {
stringstream ss;
ss << c;
cout << setw(15) << ss.str() << " -> ";
cout << linearCombo(c) << '\n';
}
 
return 0;
}</lang>
{{out}}
<pre> [1, 2, 3] -> e(1) + 2*e(2) + 3*e(3)
[0, 1, 2, 3] -> e(2) + 2*e(3) + 3*e(4)
[1, 0, 3, 4] -> e(1) + 3*e(3) + 4*e(4)
[1, 2, 0] -> e(1) + 2*e(2)
[0, 0, 0] -> 0
[0] -> 0
[1, 1, 1] -> e(1) + e(2) + e(3)
[-1, -1, -1] -> -e(1) - e(2) - e(3)
[-1, -2, 0, -3] -> -e(1) - 2*e(2) - 3*e(4)
[-1] -> -e(1)</pre>
 
=={{header|D}}==
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