实验4－1　函数的应用的答案

:π=16arctan(1/5)-4 arctan(1/239)求π。Р其中: arctan(x)=x/1-x3/3+ x5/5+‥‥+(-1)nx2n+1/(2n+1)+ ‥‥直到级数某项绝对值不大于10-15为止;π和x均为double型。Р#include<iostream.h>Рvoid main( )Р{ double a,b;Р double arctan(double x);Р a=16.0*arctan(1/5.0);Р b=4.0*arctan(1/239.0);Р cout<<"PI="<<a-b<<endl;Р}Рdouble arctan(double x)Р{ int i;Р double r,e,f,sqr;Р sqr=x*x;Р r=0; e=x; i=1;Р while(e/i>1e-15)Р { f=e/i;Р r=(i%4==1)? r+f : r-f;Р e=e*sqr; i+=2;Р }Р return r;Р}Р⒌寻找并输出11~999之间的数m,它满足m、m2和m3均为回文数。Р注:回文:各位数字左右对称的整数。Р Р例如:11满足上述条件Р 112=121,113=1331。Р#include <iostream.h>Рvoid main( )Р{ bool symm(long n);Р long m;Р for(m=11;m<1000; m++)Р if (symm(m)&&symm(m*m)&&symm(m*m*m)) Р cout<<"m="<<m<<" m*m="<<m*m<<" m*m*m="<<m*m*m<<endl;Р}Рbool symm(long n)Р{ long i, m;Р i=n; m=0;Р while(i)Р {m=m*10+i%10;Р i=i/10;Р }Р return ( m==n );Р}