摘要：
/*****x/2=1+1!/3+2!/3*5+3!/3*5*7+4!/3*5*7*9+...+i!/3*5*...(2*i+1)*****//***求x得值***/#includedouble fact(double n);double multi(double n);int main(void){ double i; double sum,item,eps; eps=1E-6; sum=1; item=1; for(i=1;item#includeint main(void){ double a,b,c; double area,perime... 阅读全文