四舍五入进行精度估算小程序

double parse(double a,int precison) //precison表示保留小数点后位数 { double A = pow(10,(double)precison); double B = pow(10,(double)(precison+1)); int T = (int)(a*A)*10+5; int U = (int)(a*B); if(U >= T) { return ((int)(a*A)+1)/A; } else return ((int)(a*A))/A; }