题解 | 时间转换
时间转换
https://www.nowcoder.com/practice/c4ae7bcac7f9491b8be82ee516a94899
#include <stdio.h> int main() { int a, b, c, d; int aa = 0xa; int bb = 0; float cc = 0.0; // scanf("%d %d ", &a, &b); // scanf(" %f" , &cc); // scanf(" %d %d", &a, &b); scanf(" %d", &a); // printf("%d %d\n", a/b,a%b ); // int date [7] = {1, 2, 3, 4, 5, 6, 7}; // int num = b >= 7 ? b % 7 : b; b = a / 3600; //小时数 c = ( a - (b * 3600) ) / 60 ; //分钟数 d = (a - (b * 3600 + c * 60 )); //秒数 // printf("%d\n", a > 7 ? a % 7 : a); printf("%d %d %d", b, c, d); return 0; }