Educational Codeforces Round 7--B. The Time

最新推荐文章于 2025-08-09 09:27:34 发布

aozil_yang

最新推荐文章于 2025-08-09 09:27:34 发布

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分类专栏： CodeForces 文章标签： c语言 codeforces

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本文介绍了一种解决24小时格式时间计算问题的方法。给定当前时间和一个分钟数，文章详细阐述了如何准确计算出经过该分钟数后的新时间，并提供了完整的C++实现代码。

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You are given the current time in 24-hour format hh:mm. Find and print the time after a minutes.

Note that you should find only the time after a minutes, see the examples to clarify the problem statement.

You can read more about 24-hour format here https://en.wikipedia.org/wiki/24-hour_clock.

Input

The first line contains the current time in the format hh:mm (0 ≤ hh < 24, 0 ≤ mm < 60). The hours and the minutes are given with two digits (the hours or the minutes less than 10 are given with the leading zeroes).

The second line contains integer a (0 ≤ a ≤ 104) — the number of the minutes passed.

Output

The only line should contain the time after a minutes in the format described in the input. Note that you should print exactly two digits for the hours and the minutes (add leading zeroes to the numbers if needed).

See the examples to check the input/output format.

        Sample test(s)
      

          input
        
23:59
10

          output
        
00:09

          input
        
20:20
121

          output
        
22:21

          input
        
10:10
0

          output
        
10:10

         大体题意：
        

         给你现在的时间，在给你一个分钟数n，求现在的时间过上n分钟后的时间。
        

         思路很清晰：
        

         1.把分钟数n先变成   小时：分钟  的格式。
        

         2.在对应和现在的时间相加。分钟满60进1，小时对24取余即可。
        

         
 
        

         代码如下：
        

         #include<bits/stdc++.h>
#define mem(x) memset(x,0,sizeof(x));
#define mem1(x) memset(x,-1,sizeof(x));
using namespace std;
const double eps = 1e-8;
const int INF = 1e8;
const int maxn = 10000 + 10;
const int maxt = 100 + 10;
typedef long long ll;
typedef unsigned long long llu;
int main()
{
    int h,m;
    while(scanf("%d:%d",&h,&m) == 2){
        int mm;
        cin >> mm;
        int lef_h = mm/60;
        int lef_m = mm-(mm/60)*60;
        int flag = 0;
        int res_m,res_h;
        res_m=m+lef_m;
        if (res_m>=60){
            flag=res_m/60;
            res_m=res_m%60;
        }
        res_h=(h+lef_h+flag)%24;
        cout << res_h/10 << res_h%10 << ":"<<res_m/10 << res_m%10 << endl;
    }
    return 0;
}
         
 
         
 
        