Educational Codeforces Round 7--B. The Time

B. The Time

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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;
}