Hi Codeforces. Today i saw [user:GreenGrape,2018-02-09] solution of the problem [D. Robot Vacuum Cleaner](http://codeforces.com/problemset/problem/922/D) : [solution](http://codeforces.com/contest/922/submission/ubmission:35053688)]. And i find there :↵
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#define debug(...) fprintf(stderr, __VA_ARGS__), fflush(stderr)↵
↵
while (t--) {↵
clock_t z = clock();↵
solve();↵
debug("Total Time: %.3f\n", (double)(clock() - z) / CLOCKS_PER_SEC);↵
}↵
~~~~~↵
↵
With this code he gets execution time of function `solve()`. `clock()` gives you current time. I have wrote helpful directivet is useful to avoid time limits. I have wrote helpful macro and decided to share with you :↵
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#define debug(...) fprintf(stderr, __VA_ARGS__), fflush(stderr)↵
#ifndef time__ ↵
#define time__(data)for (long blockTime = =NULL; (blockTime == NULL ? ?(blockTime = =clock()) != NULL : :false); debug("%s : :%.4fs" ,data ,(double) (clock() - -blockTime) / /CLOCKS_PER_SEC))↵
#endif↵
~~~~~↵
~~~~~↵
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with this u can get execution time of parts of your programm. Example:↵
↵
~~~~~↵
int main () {↵
time__("dfs time"){↵
/* dfs code */↵
}↵
}↵
~~~~~↵
And programm will print something like `dfs time : 0.265s`. You don't need to erase debugging part when you will submit it. Example submission: [submission:35088800]. ↵
Thank you for reading and I hope it will help you↵
↵
↵
~~~~~↵
#define debug(...) fprintf(stderr, __VA_ARGS__), fflush(stderr)↵
↵
while (t--) {↵
clock_t z = clock();↵
solve();↵
debug("Total Time: %.3f\n", (double)(clock() - z) / CLOCKS_PER_SEC);↵
}↵
~~~~~↵
↵
With this code he gets execution time of function `solve()`. `clock()` gives you current time. I
↵
~~~~~↵
#define debug(...) fprintf(stderr, __VA_ARGS__), fflush(stderr)↵
#define time__(d
~~~~~
~~~~~↵
↵
with this u can get execution time of parts of your programm. Example:↵
↵
~~~~~↵
int main () {↵
time__("dfs time"){↵
/* dfs code */↵
}↵
}↵
~~~~~↵
And programm will print something like `dfs time : 0.265s`. You don't need to erase debugging part when you will submit it. Example submission: [submission:35088800]. ↵
Thank you for reading and I hope it will help you↵
↵
