Algorithm

E. The Unbearable Lightness of Weights
time limit per test
1 second
memory limit per test
256 megabytes
Problemm link- https://codeforces.com/contest/1079/problem/E
input
standard input
output
standard output

You have a set of n� weights. You know that their masses are a1�1, a2�2, ..., an�� grams, but you don't know which of them has which mass. You can't distinguish the weights.

However, your friend does know the mass of each weight. You can ask your friend to give you exactly k� weights with the total mass m� (both parameters k� and m� are chosen by you), and your friend will point to any valid subset of weights, if it is possible.

You are allowed to make this query only once. Find the maximum possible number of weights you can reveal after this query.

Input

The first line contains a single integer n� (1≤n≤1001≤�≤100) — the number of weights.

The second line contains n� integers a1,a2,…,an�1,�2,…,�� (1≤ai≤1001≤��≤100) — the masses of the weights.

Output

Print the maximum number of weights you can learn the masses for after making a single query.

Examples
input
Copy
4
1 4 2 2
output
Copy
2
input
Copy
6
1 2 4 4 4 9
output
Copy
2
Note

In the first example we can ask for a subset of two weights with total mass being equal to 44, and the only option is to get {2,2}{2,2}.

Another way to obtain the same result is to ask for a subset of two weights with the total mass of 55 and get {1,4}{1,4}. It is easy to see that the two remaining weights have mass of 22 grams each.

In the second example we can ask for a subset of two weights with total mass being 88, and the only answer is {4,4}{4,4}. We can prove it is not possible to learn masses for three weights in one query, but we won't put the proof here.

 



 

Code Examples

#1 Code Example with C++ Programming

Code - C++ Programming

#include <bits/stdc++.h>

using namespace std;

int const N = 1e2 + 1;
int n, a[N], fr[N];
short dp[N][N][N*N];

short rec(int idx, int rem, int sum) {
  if(rem < 0 || sum < 0)
    return 0;
  if(idx == N)
    return rem == 0 && sum == 0;

  short &ret = dp[idx][rem][sum];
  if(ret != -1)
    return ret;
  ret = min(2, int(rec(idx + 1, rem, sum)));

  for(int i = 1; i <= fr[idx]; ++i)
    ret = min(2, ret + rec(idx + 1, rem - i, sum - idx * i));

  return ret;
}

int main() {
  scanf("%d", &n);
  for(int i = 0; i < n; ++i)
    scanf("%d", a + i), ++fr[a[i]];

  int cnt = 0;
  for(int i = 0; i < N; ++i)
    cnt += fr[i] != 0;

  if(cnt <= 2) {
    printf("%d\n", n);
    return 0;
  }

  memset(dp, -1, sizeof dp);
  int res = 0;
  for(int i = 1; i < N; ++i)
    for(int j = 1; j <= fr[i]; ++j)
      if(rec(1, j, i * j) == 1)
        res = max(res, j);
  printf("%d\n", res);

  return 0;
}
Copy The Code & Try With Live Editor

Input

x
–
+
cmd
4 1 4 2 2

Output

x
–
+
cmd
2
Advertisements

Demonstration

Codeforces Solution-E. The Unbearable Lightness of Weights-Solution in C, C++, Java, Python,The Unbearable Lightness of Weights,Codeforces Solution

Previous
Codeforces solution 1080-B-B. Margarite and the best present codeforces solution
Next
CodeChef solution DETSCORE - Determine the Score CodeChef solution C,C+