Google Code Jam Minimum Scalar Product - c++

Question

I was solving a problem from Google Code Jam and I am not able to solve the problem: http://code.google.com/codejam/contest/32016/dashboard#s=p0 (Minimum Scalar Product, Problem A 2008)
The strategy I used was:
Accept v1 and v2 from the user
Sort both v1 and v2
Reverse v2 i.e. sort v2 in descending order
Multiply straight-out corresponding v1[i] * v2[i] and store the result in product
Sum up all such products and print the answer
I did some research and indeed it appears that's the only permutation that's possible to obtain. However, my code does not produce the correct output:
#include <stdio.h>
#include <algorithm>
#include <iostream>
using namespace std;
int main()
{
int T;
int cases;
FILE *fin = fopen ("A-small-practice.in", "r"); // open input file
FILE *fout = fopen ("output.out", "w");
fscanf(fin, "%d", &T);
for(cases = 1; cases <= T; cases++)
{
int v1[1000], v2[1000];
int i,j; int n;
int product =0;
fscanf(fin, "%d", &n);
for(i=0; i < n; i++)
{
fscanf(fin, "%d",&v1[i]);
fscanf(fin, "%d", &v2[i]);
}
sort(v1,v1+n);
sort(v2,v2+n);
reverse(v1, v1+n);
int k;
for(k = 0; k < n; k++)
{
product += v1[k] * v2[k];
}
fprintf(fout, "Case #%d: %d\n", cases, product);
}
return 0;
}

Answer 1

You should use long long.
This worked for me:
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
long long T,n,v1[1000],v2[1000];
cin >> T;
for (int t = 1; t <= T; t++) {
cin >> n;
for (int i = 0; i < n; i++)
cin >> v1[i];
for (int i = 0; i < n; i++)
cin >> v2[i];
sort(v1,v1+n);
sort(v2,v2+n);
long long p = 0;
for (int i = 0; i < n; i++)
p += v1[i]*v2[n-i-1];
cout << "Case #" << t << ": " << p << endl;
}
return 0;
}

Answer 2

After hours of troubleshooting, I found that the problem was in the way input was taken, as I had used only one loop for both v1 & v2:
for(i=0; i < n; i++)
{
fscanf(fin, "%d",&v1[i]);
fscanf(fin, "%d", &v2[i]);
}
And it should have been done this way:
for(i=0; i<n; i++)
fscanf(fin, "%d",&v1[i]);
for(i=0; i<n; i++)
fscanf(fin, "%d", &v2[i]);
Thank you lukas1994 and pkacprzak

Answer 3

This worked for me:
#include <algorithm>
#include <iostream>
#include <vector>
#include <numeric>
using std::vector;
long long min_dot_product(size_t n, vector<int> a, vector<int> b) {
long long result = 0;
if (n != 0)
{
std::sort(a.begin(), a.end());
std::sort(b.begin(), b.end());
std::reverse(a.begin(), a.end());
/*for (size_t i = 0; i < n; i++) {
result += a[i] * b[n - 1 - i];
}*/
result = std::inner_product(a.begin(), a.end(), b.begin(), 0);
}
return result;
}
int main() {
size_t n;
std::cin >> n;
vector<int> a(n), b(n);
for (size_t i = 0; i < n; i++) {
std::cin >> a[i];
}
for (size_t i = 0; i < n; i++) {
std::cin >> b[i];
}
std::cout << min_dot_product(n, a, b) << std::endl;
}

Answer 4

There is a specific case to be considered.
When one vector is all negative and the other all positive, the algorithm does not produce the minimum result.
if x = [-1,-2] and x = [1,2]
the minimum according to the algorithm output is :
(-1*2) + (-2*1) = -4
However, if you instead use x1y1 + x2y2... xnyn for this case you get:
(-1 * 1) + ( -2*2) = -1-4 = -5