c语言求方程ax2 bx c=0,用三个函数
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#includedoublefunc(doublex){returnx*(x-3.0)-5.0;}doublebinary_search(doublex1,doublex2,double(*f)(do
#include#includevoidmain(){floatc,d,f,a,b;a=-10;b=10;while(fabs(a-b)>=0.000001){c=(a+b)/2;d=2*a*a*a-
#includevoidmain(){floats,f0,h,x;intn,i;printf("inputn:");scanf("%d",&n);h=1.0/n;f0=4.0;s=0.0;for(i=
#include#includevoidmain(){floata,b,c;printf("inputa,b,c:");scanf("%f%f%f",&a,&b,&c);if(b*b-4*a*c>=0
#include <stdio.h>#include <assert.h>double f(double x){
这道题其实很简单的,就只是三个if而已吧,你就根据求根公式编写就行.可能求虚根难点.这些要求和提示已经很详细了,除非你没学过c语言.再问:我不是不懂,是a,b,c,x1,x2,disc=b*b-4*a
#include#includemain(){doublea,b,c,x1,x2;printf("Inputa,b,c:\n");scanf("%lf%lf%lf",&a,&b,&c);{if(a=0
#include#include"math.h"usingnamespacestd;doublefangcheng(doublea,doubleb,doublec){if(b*b-4*a*c>=0)
第一个地方:doubley1=(-b+sqrt(b*b-4*a*c))/(2*a);doubley2=(-b-sqrt(b*b-4*a*c))/(2*a);printf("x的值为%lf或者%lf",
#include#includeintmain(){floata,b,c;floatx1,x2;scanf("%f%f%f",&a,&b,&c);if(b*b-4*a*c
#include#include#defineDeltab*b-4*a*cintmain(){floatx1,x2;inta,b,c;scanf("%d%d%d",&a,&b,&c);x1=(-b+(
z=(y-x)/2.0;把这句的减号变为加号.求中点是(x+y)/2
设带表头结点的双向链表的定义为typedefintElemTyp*:typedefstructdnode{file://双向链表结点定义ElemTypedata:file://数据structdnod
#include#includedoubleeps=10E-6;doublef(doublek)//原函数方程{returnlog10(k)+k-2.0;}doubleget(doublek){ret
intf(floatx)返回值,错了.
#includmain(){inta,b,c;floatm,n,k,l,p,q;chars;s='i';printf("inputabc:/n");scanf("%d%d%d",&a,&b,&c);i
很简单的#include#includevoidf1(doublea,doubleb,doublec){cout
0x80这是十六进制数,变成十进制数为-128,因为char型在C语言中范围为-128~127,并不是0乘以80,c语言中乘以用*,例如0*80,表示0乘以80;
#include"stdio.h"#include"math.h"main(){doublea,b,c,x1,x2,dt,p,q;printf("pleaseinputa,b,c:");scanf("
#include#include#defineeps5e-6#definedelta1e-6#definee2.718floatbisection(floata,floatb,float(*f)(fl