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  1. #include <iostream>
  2. using namespace std;
  3.  
  4. int main() {
  5.     int T;
  6.     cin >> T;
  7.     while (T--) {
  8.         long long L, R;
  9.         cin >> L >> R;
  10.         // 计算区间[L, R]中奇数的个数
  11.         long long count = (R - L + 2) / 2;
  12.         if ((L % 2 == 0) && (R % 2 == 0)) {
  13.             count = (R - L) / 2;
  14.         }
  15.         cout << count << endl;
  16.     }
  17.     return 0;
  18. }
Success #stdin #stdout 0s 5320KB
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stdin
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stdout
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https://ideone.com/SEouDQ
problem DHRODDMX, based on SPOJ submission 34546498 (CPP14), 2025-05-06 03:14:58
language:
C++14 (gcc 8.3)
created:
1 day ago
visibility:
public

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