Codeforces Round #320 (Div. 1) [Bayan Thanks-round] a problem about Polyline

The functions given in the topic are periodic, can always be moved to the first period, of course, a<b is no solution.

Assuming that the ascending segment is moved, then there is a-2*x*n=b, and note that the limiting condition X≥b,n is an integer, then n≤ (A-B)/(2*B). X≥ (A-B)/(2*N) satisfying the condition

Assumed in the descending segment, 2*x-(a-2*x*n) =b,n+1≤ (a+b)/(2*b), x≥ (a+b)/(n+1)

Take the minimum value of both

#include <bits/stdc++.h>using namespacestd;intMain () {intb; scanf"%d%d",&a,&b); if(A<b) puts ("-1"); Else {        intT1 = a+b, t2 = A-b; intN1 = t1/(2*B) *2, N2 = t2/(2*B) *2; DoubleA1 = T1*1./n1, a2 = t2*1./N2; printf ("%.10lf\n", Min (a1,a2)); }    return 0;}

Codeforces Round #320 (Div. 1) [Bayan Thanks-round] a problem about Polyline