C++ 实现：十个数奇偶位置累加值相减的计算程序

```cpp #include <iostream> using namespace std; int main() { int nums[10]; int sum1 = 0, sum2 = 0; for (int i = 0; i < 10; i++) { cin >> nums[i]; } for (int i = 0; i < 10; i++) { if (i % 2 == 0) { sum2 += nums[i]; } else { sum1 += nums[i]; } } int result = sum1 - sum2; cout << result << endl; return 0; } ``` ######[AI问答 | 135点数解答 | 2025-12-07 14:35:12]