我正在解决代码强制问题，代码功能没有问题，但是代码超出了内存使用量。有人可以解释如何改善这一点吗？编程时通常如何学习内存管理，因为我没有发现任何与此相关的东西。这是我的代码：
问题摘要：给您6个输入数字，前5个数字应分别乘以另一个整数，并且相乘后的总和为输入中的sixth整数。您应该找到可以与输入中的每个值相乘的所有数字组合，以求和。输出基本上是要乘以输入中每个数字所选择的整数的总和。
#include <iostream>
#include <vector>
#ifndef MAX
#define MAX 100
#endif
using namespace std;
void storeAndFilter(vector <int> &arr,int chosenNumConst, int mypasha);
void print(vector<int> &v);
int printsum(vector<int> &v);
int main(int argc, char const *argv[])
{
//array of input options
int a1, a2, a3, a4, a5, pasha;
cin >> a1 >> a2 >> a3 >> a4 >> a5 >> pasha;
//declarations of vectors
vector<int> arrStrA1;
vector<int> arrStrA2;
vector<int> arrStrA3;
vector<int> arrStrA4;
vector<int> arrStrA5;
//sorting and filtering the vectors
storeAndFilter(arrStrA1,a1,pasha);
storeAndFilter(arrStrA2,a2,pasha);
storeAndFilter(arrStrA3,a3,pasha);
storeAndFilter(arrStrA4,a4,pasha);
storeAndFilter(arrStrA5,a5,pasha);
//cout<<"All Posibilities valid (Minimized by removing values < pasha) : "<<endl;
// print (arrStrA1);
// print (arrStrA2);
// print (arrStrA3);
// print (arrStrA4);
// print (arrStrA5);
//scores vectors
vector<int> resultsA1;
vector<int> resultsA2;
vector<int> resultsA3;
vector<int> resultsA4;
vector<int> resultsA5;
int i,j,k,l,m;
for (i=0; i < (int)arrStrA1.size(); ++i)
{
for (j=0; j < (int)arrStrA2.size(); ++j)
{
for (k=0; k < (int)arrStrA3.size(); ++k)
{
for (l=0; l < (int)arrStrA4.size(); ++l)
{
for (m=0; m < (int)arrStrA5.size(); ++m)
{
if(arrStrA1.at(i)+arrStrA2.at(j)+arrStrA3.at(k)+arrStrA4.at(l)+arrStrA5.at(m)==pasha)
{
resultsA1.push_back(arrStrA1.at(i));
resultsA2.push_back(arrStrA2.at(j));
resultsA3.push_back(arrStrA3.at(k));
resultsA4.push_back(arrStrA4.at(l));
resultsA5.push_back(arrStrA5.at(m));
}
}
}
}
}
}
//divise each term by the card value
for (int i = 0; i < (int)resultsA1.size(); ++i)
{
if (a1==0)
resultsA1.at(i) /= 1;
else
resultsA1.at(i) /= a1;
}
for (int i = 0; i < (int)resultsA2.size(); ++i)
{
if (a2==0)
resultsA2.at(i) /= 1;
else
resultsA2.at(i) /= a2;
}
for (int i = 0; i < (int)resultsA3.size(); ++i)
{
if(a3==0)
resultsA3.at(i) /= 1;
else
resultsA3.at(i) /= a3;
}
for (int i = 0; i < (int)resultsA4.size(); ++i)
{
if (a4==0)
resultsA4.at(i) /= 1;
else
resultsA4.at(i) /= a4;
}
for (int i = 0; i < (int)resultsA5.size(); ++i)
{
if(a5==0)
resultsA5.at(i) /= 1;
else
resultsA5.at(i) /= a5;
}
//Uncomment to show the table list after division
// print(resultsA1);
// print(resultsA2);
// print(resultsA3);
// print(resultsA4);
// print(resultsA5);
int scra1=printsum(resultsA1);
int scra2=printsum(resultsA2);
int scra3=printsum(resultsA3);
int scra4=printsum(resultsA4);
int scra5=printsum(resultsA5);
cout << scra1 <<" "<< scra2 <<" "<< scra3 <<" "<<scra4 <<" "<< scra5 <<endl;
return 0;
}
void print(vector<int> &v)
{
int size = v.size();
cout<<"========================"<<endl;
for (int i = 0; i < size; ++i)
cout<<v.at(i)<<endl;
cout<<"========================"<<endl;
}
int printsum(vector<int> &v)
{
int sum =0;
for (int i = 0; i < (int)v.size(); ++i)
sum += v.at(i);
return sum;
}
void storeAndFilter(vector <int> &arr,int chosenNumConst, int mypasha)
{
arr.reserve(10);
int i=0;
for (; i <= MAX; ++i)
{
arr.push_back(i*chosenNumConst);
if (arr.at(i)>mypasha)
break;
}
arr.resize(i);
}
我想到的一些东西：
但是要清楚，嵌套的for循环并没有进行太多的计算，他们发现5个数字的全部组合'5个循环'求和成一个特定值。在进入循环之前应用了过滤，因此嵌套循环可能不是问题。问题中的最大内存限制是：256 MB
通过不使用所有这些向量，您可以使用更少的内存。您可以这样编写代码：
// make sure we handle zero properly
auto end = [&](int num){
return num == 0 ? num : pasha/num;
};
for (auto i=0, end_i = end(a1); i <= end_i; ++i)
{
for (auto j=0, end_j = end(a2); j <= end_j; ++j)
{
for (auto k=0, end_k = end(a3); k <= end_k; ++k)
{
for (auto l=0, end_l = end(a4); l <= end_l; ++l)
{
for (auto m=0, end_m = end(a5); m <= end_m; ++m)
{
if(a1*i+a2*j+a3*k+a4*l+a5*m==pasha)
{
std::cout << i << " " << j << " " << k << " " << l << " " << m << "\n";
}
}
}
}
}
}
并且它输出所有有效结果。例如，输入1 2 3 4 5 6会产生
0 0 2 0 0
0 1 0 1 0
0 3 0 0 0
1 0 0 0 1
1 1 1 0 0
2 0 0 1 0
2 2 0 0 0
3 0 1 0 0
4 1 0 0 0
6 0 0 0 0
请参见工作示例here