2 solutions
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1
#include <iostream> using namespace std; int main() { int n; cin >> n; long long product = 1; // 使用long long防止溢出 const long long LIMIT = 1000000; for (int i = 0; i < n; i++) { int a; cin >> a; // 如果当前乘积已经超过或等于LIMIT,保持不变即可 if (product <= LIMIT) { product *= a; } } if (product > LIMIT) { cout << ">1000000" << endl; } else { cout << product << endl; } return 0; }
Information
- ID
- 2164
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- 5
- Tags
- # Submissions
- 39
- Accepted
- 16
- Uploaded By