CodeForces - 1692D
Victor has a 24-hour clock that shows the time in the format "HH:MM" (00 ≤≤ HH ≤≤ 23, 00 ≤≤ MM ≤≤ 59). He looks at the clock every xx minutes, and the clock is currently showing time ss.
How many different palindromes will Victor see in total after looking at the clock every xx minutes, the first time being at time ss?
For example, if the clock starts out as 03:12 and Victor looks at the clock every 360360 minutes (i.e. every 66 hours), then he will see the times 03:12, 09:12, 15:12, 21:12, 03:12, and the times will continue to repeat. Here the time 21:12 is the only palindrome he will ever see, so the answer is 11.
A palindrome is a string that reads the same backward as forward. For example, the times 12:21, 05:50, 11:11 are palindromes but 13:13, 22:10, 02:22 are not.
Input
The first line of the input contains an integer tt (1≤t≤1001≤t≤100) — the number of test cases. The description of each test case follows.
The only line of each test case contains a string ss of length 55 with the format "HH:MM" where "HH" is from "00" to "23" and "MM" is from "00" to "59" (both "HH" and "MM" have exactly two digits) and an integer xx (1≤x≤14401≤x≤1440) — the number of minutes Victor takes to look again at the clock.
Output
For each test case, output a single integer — the number of different palindromes Victor will see if he looks at the clock every xx minutes starting from time ss.
Examples
| Inputcopy | Outputcopy |
|---|---|
6 03:12 360 00:00 1 13:22 2 15:15 10 11:11 1440 22:30 27 | 1 16 10 0 1 1 |
#include<bits/stdc++.h>
using namespace std;
#define int long longsigned main(){int t;cin>>t;while(t--){int h,f,k;char c;string x;cin>>x>>k;//输入时间 ，每次相加分数 h=(x[0]-'0')*10+(x[1]-'0');//求小时 f=(x[3]-'0')*10+(x[4]-'0');//求分 int h1=h,f1=f;int sum=0,p=0;while((h1!=h||f1!=f)||(p==0)){//循环直到时间重新与输入时间相等 p++;int q=f1+k;//求分数相加和 h1=(h1+q/60)%24,f1=q%60;//更新时间 string z=to_string(h1);//把时间转换为字符串 if(z.size()<2){//如果字符串只有一位就在前面+‘0’ z='0'+z;}string c=to_string(f1);if(c.size()<2){c='0'+c;}string ss=z+":"+c;string dd=ss;reverse(dd.begin(),dd.end());//反转字符串 if(ss==dd){//比较是否为回文串 sum++;}}cout<<sum<<endl;}}
这样可以吗