B1026 program running time

1026 Program running time (15 branch )

To get one C The running time of language program , A common method is to call the header file time.h, It provides clock() function , You can capture from the beginning of the program to clock() The time taken to be called . This time unit is clock tick, namely “ Clock strike ”. And there's also a constant CLK_TCK, The number of clock ticks per second of the machine clock is given . So in order to get a function f Running time of , We just call f Call before clock(), Get a clock beat C1; stay f Call... After the execution is completed clock(), Get another clock beat C2; The difference between the number of clock ticks obtained twice (C2-C1) Namely f The number of clock ticks consumed by the operation , Then divide by the constant CLK_TCK, You get the running time in seconds .

Let's simply assume that the constant CLK_TCK by 100. Now, given the number of clock beats obtained before and after the measured function , Please give the running time of the tested function .

Input format ：

The input is given in order on one line 2 It's an integer C1 and C2. Note that the clock points obtained twice must be different , namely C1 < C2, And the value is in [0,10 Of 7 Power ].

Output format ：

Output the running time of the function under test in one line . The running time must be in accordance with hh:mm:ss（ namely 2 Bit when : branch : second ） Format output ; Insufficient 1 The time of seconds is rounded to seconds .

sample input

123 4577973

sample output

12:42:59

Topic analysis ：

1. There's nothing to say

The code is as follows ：

#include<bits/stdc++.h>
#define CLK_TCK 100
using namespace std;
int main(){
    int a,b;
    cin>>a>>b;
    double z=1.0*(b-a)/CLK_TCK;
	int number=round(z);
	printf("%02d:%02d:%02d\n",number/3600,(number/60)%60,number%60);
    
    return 0;
}