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back to boardHow this problem can be accepted in 0.001? Posted by giorgi 4 Sep 2011 21:13 Here is my code which accepted in 0.015, can someone tell me what to optimize else to get it accepted in 0.001?(sorry if my english is not very good) #include <cstdio> #include <cmath> using namespace std; int main() { int V; double a; double K; scanf("%d %Lf %Lf", &V, &a, &K); double s=0; if(a!=0&&a!=90) { double p = a*3.1415926535/180; double Vy = V*sin(p); double Vx = V*cos(p); s = Vx*Vy*0.2/(1-1/K); } printf("%.2f",s); return 0; } Re: How this problem can be accepted in 0.001? Posted by Pegasus 21 Oct 2012 18:10 May be you can find a formula and count it directly Re: How this problem can be accepted in 0.001? Posted by Bogatyr 22 Oct 2012 03:49 > can someone tell me what to optimize else to get it accepted in 0.001? The fastest possible submit time on the current judge seems to be 0.015, so you're already at the best possible time. In earlier years the fastest time was 0.001. It's impossible to beat or even match the older solutions unless the judge's time measurement system changes. |
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