1.不能用%d输出double类型的数
double a1=5.3;
double a2=1234.1234;
double a3=3412341.12341234; double b1=1.5;
double b2=123.5;
double b3=23412.5; double c1=10.7;
double c2=2.9;
double c3=3241324.56251;
printf("%d\n",a1);
printf("%d\n",a2);
printf("%d\n",a3); printf("%d\n",b1);
printf("%d\n",b2);
printf("%d\n",b3); printf("%d\n",c1);
printf("%d\n",c2);
printf("%d\n",c3);
输出的值不会变
2.
float pow(float x, float y)
x,y可以为整形，因为进行时会自动转换类型，
但是用%d输出double类型的数则不行。
可以使用(int)。
cout<<pow(,)<<endl;
cout<<pow(,)<<endl;
cout<<pow(,)<<endl; printf("\n"); printf("%f\n",pow(,));
printf("%f\n",pow(,));
printf("%f\n",pow(,)); printf("\n"); printf("%d\n",pow(,));
printf("%d\n",pow(,));
printf("%d\n",pow(,)); printf("\n"); printf("%d\n",(int)pow(,));
printf("%d\n",(int)pow(,));
printf("%d\n",(int)pow(,));
3.
float log(float x)
(int)使用在在(int)右边的第一个整数。
所以不能使用(int)log(s)/log(2)，而是(int)(log(s)/log(2))。
而求log(2^k)/log(2)时，出现问题，因为是精度不准的问题，如结果为1.99..9xxxd，用(int)后结果为1。
可以采用log(2^k+minv)/log(2)解决，minv=1e-12(或其它)。
printf("%.15f\n",log()/log());
printf("%.15f\n",log()/log());
printf("%.15f\n",log()/log());
printf("%.15f\n",log()/log());
printf("%.15f\n",log()/log());
printf("\n");
printf("%f\n",(int)log()/log());
printf("%f\n",(int)log()/log());
printf("%f\n",(int)log()/log());
printf("%f\n",(int)log()/log());
printf("%f\n",(int)log()/log());
printf("\n");
printf("%d\n",(int)(log()/log()));
printf("%d\n",(int)(log()/log()));
printf("%d\n",(int)(log()/log()));
printf("%d\n",(int)(log()/log()));
printf("%d\n",(int)(log()/log()));
printf("%d\n",(int)(log()/log()));
printf("%d\n",(int)(log()/log()));
printf("\n");
printf("%d\n",(int)(log(+minv)/log()));
printf("%d\n",(int)(log(+minv)/log()));
printf("%d\n",(int)(log(+minv)/log()));
printf("%d\n",(int)(log(+minv)/log()));
printf("%d\n",(int)(log(+minv)/log()));
printf("%d\n",(int)(log(+minv)/log()));
printf("%d\n",(int)(log(+minv)/log()));
