1 solutions
-
0
C++ :
#include <iostream> #include <algorithm> // 用于 min 函数 using namespace std; int main() { int A, B, C; cin >> A >> B >> C; // 计算最多可以买多少个糖瓜 int max_possible = B / A; // 取 min(max_possible, C) 作为最终结果 int result = min(max_possible, C); cout << result << endl; return 0; }
- 1
Information
- ID
- 696
- Time
- 1000ms
- Memory
- 128MiB
- Difficulty
- 10
- Tags
- # Submissions
- 1
- Accepted
- 1
- Uploaded By