PAT-BASIC1026——程序运行时间

我的PAT-BASIC代码仓：https://github.com/617076674/PAT-BASIC

原题链接：https://pintia.cn/problem-sets/994805260223102976/problems/994805295203598336

题目描述：

知识点：时分秒转换

思路：按题述一步步运算即可

两个注意点：

（1）不足1秒的时间四舍五入。

（2）个位数需要前面补0。

时间复杂度是O(n / 60)，其中n为输入两个时间的时间间隔。空间复杂度是O(1)。

C++代码：

#include<iostream>
#include<time.h>

using namespace std;

int main() {
	long startTime;
	long endTime;

	cin >> startTime >> endTime;

	long time = (endTime - startTime) / 100;
	if ((endTime - startTime) % 100 >= 50) {
		time++;
	}

	if (time >= 60 * 60) {
		int hour = time / (60 * 60);
		if (hour <= 9) {
			cout << "0" << hour << ":";
		} else {
			cout << hour << ":";
		}
		int miniute = (time - hour * 60 * 60) / 60;
		if (miniute <= 9) {
			cout << "0" << miniute << ":";
		} else {
			cout << miniute << ":";
		}
		int second = time - hour * 60 * 60 - miniute * 60;
		if (second <= 9) {
			cout << "0" << second;
		} else {
			cout << second;
		}
	} else if (time >= 60) {
		cout << "00:";
		int miniute = time / 60;
		if (miniute <= 9) {
			cout << "0" << miniute << ":";
		} else {
			cout << miniute << ":";
		}
		int second = time - miniute * 60;
		if (second <= 9) {
			cout << "0" << second;
		} else {
			cout << second;
		}
	} else {
		cout << "00:00:";
		int second = time;
		if (second <= 9) {
			cout << "0" << second;
		} else {
			cout << second;
		}
	}

}

C++解题报告：