C++ – T84891 Part

Question link:https://www.luogu.org/problem/T84891

topic

Ideas

If part A conflicts with part B, and part A conflicts with part C, then part B and part C also conflict.

That is to say, among part A, part B, and part C, we can only select one part.

Part A, Part B, and Part C have a common feature, that is, the same value after %K.

So just put all the input content into a set after %K, and finally output the size of the set.

代码

#include <cstdio>
#include <algorithm>
#include <bitset>
#include <cmath>
#include <set>
using namespace std;
const int MAXN = 1e5 + 100;
int arr[MAXN];
bitset<MAXN> vis;
int main() {
    vis.reset();
    int n, k;
    scanf("%d %d", &n, &k);
    for(int i = 0; i < n; i++) 
        scanf("%d", &arr[i]);
    //按照惯例，暴力做法注释
//  sort(arr, arr + n);
//  for(int i = 0; i < n; i++) {
//      for(int j = i + 1; j < n; j++) {
//          if(!vis[j]) {
//              if((arr[j] - arr[i]) % k == 0)
//                  vis[j] = true;
//          }
//      }
//  }
    set<int> s;
    for(int i = 0; i < n; i++)
        s.insert(arr[i] % k);
//  printf("%d\n", n - vis.count());
    printf("%d\n", s.size());
    return 0;
}