题解 | #时间转换#
时间转换
http://www.nowcoder.com/practice/c4ae7bcac7f9491b8be82ee516a94899
利用数组存储，并反向输出
#include using namespace std; int main() { int a, arr[3] = { 0 }, i; cin >> a; for ( i = 0; i < 3; i++) { arr[i] = a % 60; a /= 60; } for (i--; i >= 0; i--) cout << arr[i] << " "; system("pause"); return 0; }
