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46 lines
2.1 KiB
46 lines
2.1 KiB
/*
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第一个的文字：小A的寝室里放着一个时钟。时钟有时针，分针，秒针三种指针。在某些时刻，时钟会记录下当前时间，格式为hh:mm:ss。
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时钟上看不出日期，每天零点，三根指针都会归零。现在小A得到了一系列被记录的时间，且这些时间是按照先后顺序被记录的。
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小A想让你算算，每两个时间点之间，秒针至少转了多少圈。注:一天有 24 个小时，1 小时有 60 分钟，1分钟有 60 秒，秒针一分钟转一圈。
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输入描述：第一行一个整数 n(2 ≤n < 10e5)，代表时间序列中时间点的个数。第二行n个字符串，每个字符串代表一个时间点，格式为hh:mm:ss(0 ≤ hh< 24, 0≤ mm< 60, 0 ≤ ss < 60).
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输出描述:n -1个数，第i个数表示第i到i十1时间点秒针转过的圈数。输出结果和答案的绝对误差或相对误差不超过 10e-5 即被认为正确。
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*/
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#include <iostream>
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#include <vector>
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#include <string>
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#include <cmath>
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#include <iomanip>
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using namespace std;
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// Helper function to convert time in "hh:mm:ss" format to total seconds from midnight
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int timeToSeconds(const string& time) {
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int hh = stoi(time.substr(0, 2));
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int mm = stoi(time.substr(3, 2));
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int ss = stoi(time.substr(6, 2));
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return hh * 3600 + mm * 60 + ss;
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}
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int main() {
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int n;
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cin >> n;
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vector<string> times(n);
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// Input all time points
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for (int i = 0; i < n; ++i) {
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cin >> times[i];
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}
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// Calculate and print the number of full rotations of the second hand between consecutive time points
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for (int i = 1; i < n; ++i) {
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int t1 = timeToSeconds(times[i - 1]);
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int t2 = timeToSeconds(times[i]);
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// If t2 is smaller, it means the time passed to the next day
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if (t2 < t1) {
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t2 += 24 * 3600; // add 24 hours in seconds
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}
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int secondsDiff = t2 - t1;
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double rotations = secondsDiff / 60.0; // calculate the full rotations
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printf("%.*f", 10, rotations);
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// cout << setprecision(10) << rotations << endl; // print the result
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}
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return 0;
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}
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