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  1. #include <iostream>
  2. #include <vector>
  3. #include <unordered_map>
  4. #include <algorithm>
  5. using namespace std;
  6.  
  7. const int MAX_A = 1000000;
  8.  
  9. int main() {
  10.     ios_base::sync_with_stdio(false);
  11.     cin.tie(nullptr);
  12.  
  13.     // 筛法求最小质因子
  14.     vector<int> minPrime(MAX_A + 1, 0);
  15.     for (int i = 2; i <= MAX_A; i++) {
  16.         if (minPrime[i] == 0) {
  17.             minPrime[i] = i;
  18.             for (long long j = (long long)i * i; j <= MAX_A; j += i) {
  19.                 if (minPrime[j] == 0) {
  20.                     minPrime[j] = i;
  21.                 }
  22.             }
  23.         }
  24.     }
  25.  
  26.     int n;
  27.     cin >> n;
  28.  
  29.     unordered_map<int, int> firstOccur;
  30.     firstOccur[0] = 0; // 初始状态（空乘积）在位置0
  31.  
  32.     int state = 0;
  33.     int maxLen = -1;
  34.  
  35.     for (int i = 1; i <= n; i++) {
  36.         int x;
  37.         cin >> x;
  38.  
  39.         // 分解质因数，更新状态
  40.         while (x > 1) {
  41.             int p = minPrime[x];
  42.             int cnt = 0;
  43.             while (x % p == 0) {
  44.                 x /= p;
  45.                 cnt++;
  46.             }
  47.             if (cnt % 2 == 1) {
  48.                 // 奇数次出现，翻转该质因子的奇偶性
  49.                 state ^= p;
  50.             }
  51.         }
  52.  
  53.         // 检查之前是否出现过相同状态
  54.         if (firstOccur.count(state)) {
  55.             int len = i - firstOccur[state];
  56.             if (len > maxLen) {
  57.                 maxLen = len;
  58.             }
  59.         } else {
  60.             firstOccur[state] = i;
  61.         }
  62.     }
  63.  
  64.     if (maxLen <= 0) {
  65.         cout << -1 << '\n';
  66.     } else {
  67.         cout << maxLen << '\n';
  68.     }
  69.  
  70.     return 0;
  71. }
Success #stdin #stdout 0.01s 7044KB
comments (?)
stdin 
4
2 3 5 7
stdout 
-1
https://ideone.com/nrjrct
language:
C++ (gcc 8.3)
created:
2 days ago
visibility:
public

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