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Codeforces Round #310 (Div. 1) D. Case of a Top Secret 二分 stl应用(二)
xr; yl = xr - (yr - xr); it = myset.lower_bound(make_pair(yl,-1)); xl = x[it->second]; l2 = xl - yl; if(l2 == l){ out(it->second); putchar(' '); break; } else { if(lxl == xl && lxr == xr){ l = l % (l - l2); } else { l = l2; } s = xl; lxl = xl;lxr = xr; } } } } return 0; }