c语言玫瑰花代码
#include <iostream>
#include <cmath>
using namespace std;
void getRoseNum(int lower,int upper);
bool isRoseNum(int n);
void main()
{
int upper,lower;
cout<<"请输入下界:"<<endl;
cin>>lower;
cout<<"请输入上界:"<<endl;
cin>>upper;
cout<<"所有玫瑰花数:"
getRoseNum(lower,upper);
}
void getRoseNum(int lower,int upper)
{
if((lower<1000)||(upper>9999))
{
cout<<"上下界错误!"<<endl; return;
}
for (int i=lower;i<=upper;i++)
{
if (isRoseNum(i))
{
cout<<i<<endl;
}
}
}
bool isRoseNum(int n)
{
char a[5]={'0'};//这里改一下就行了，不然会溢出
itoa(n,a,10);
int sum=0;
for (int i=0;i<4;i++)
sum+=pow((double)(a[i]-48),4);
if (n==sum) return true;
return false;
}
㈡ c语言编写程序 水仙花数 玫瑰花数
1
#include <stdio.h>
#include <stdlib.h>
int flower(int n)
{
int i, j, k;
i = n % 10;
j = n / 10 % 10;
k = n / 100;
if (i*i*i + j*j*j + k*k*k == n)
return 1;
else
return 0;
}
int main(void)
{
int i;
for (i = 100; i < 1000; i++)
{
if (flower(i) == 1)
printf("%d ", i);
}
return 0;
}
2.
#include <stdio.h>
#include <stdlib.h>
int rose(int n)
{
int i, j, k,m;
i = n % 10;
j = n / 10 % 10;
k = n / 100%10;
m = n / 1000;
if (i*i*i*i + j*j*j*j + k*k*k*k+m*m*m*m == n)
return 1;
else
return 0;
}
int main(void)
{
int i;
for (i = 1000; i < 10000; i++)
{
if (rose(i) == 1)
printf("%d ", i);
}
return 0;
}
㈢ c语言中什么是玫瑰花数
c语言中的玫瑰花数是指一个四位数等于它的各数位上的数字的四次方和。
c语言介绍；C语言是一门通用计算机编程语言，应用广泛，C语言的设计目标是提供一种能以简易的方式编译，产生少量的机器码以及不需要任何运行环境支持便能运行的编程语言。
㈣ C语言：一行一个，在屏幕上输出所有四叶玫瑰数，用while循环，咋编
按照你的要求编写的输出所有四叶玫瑰数的C语言程序如下
㈤ 你是不是有999玫瑰C语言版源代码能给我吗
#include <dos.h>
#include <graphics.h>
#include <math.h>
/*玫瑰花*/
#define FNX(x) (int)(xo+(x)*1.0)
#define FNY(y) (int)(getmaxy()-(yo+(y)*1.0))
#define FNX2(phi) cos(phi)*ac-sin(phi)*bs
#define FNY2(phi) cos(phi)*as+sin(phi)*bc
/*画旋转的椭圆*/
void elli(int xo,int yo,int a,int b,double theta)
{
int i;
double da,c,s,ac,as,bc,bs,xf,yf,phi,x,y;
theta=theta*0.01745;
da=3*0.1745;
c=cos(theta);
s=sin(theta);
ac=a*c;as=a*s;
bc=b*c;bs=b*s;
x=FNX2(0);
y=FNY2(0);
moveto(FNX(x),FNY(y));
for(i=1;i<=360;i++)
{
phi=i*da;
xf=x*cos(phi)*0.1;
yf=b*sin(phi)*0.1;
x=FNX2(phi);
y=FNY2(phi);
lineto(FNX(x),FNY(y));
}
}
/*花*/
void hua(int x,int y)
{
register i;
/*画粉红色玫瑰*/
setcolor(12);
arc(x+65,y-60,150,350,8);
arc(x+66,y-54,300,470,8);
arc(x+65,y-56,30,230,10);
arc(x+64,y-57,300,490,17);
ellipse(x+73,y-30,250,450,27,40);
ellipse(x+59,y-30,100,290,27,40);
ellipse(x+65,y-40,140,270,20,30);
setfillstyle(SOLID_FILL,5);
floodfill(x+65,y-20,12);
/*画红色玫瑰*/
arc(x,y,150,350,12);
arc(x+1,y+8,280,470,12);
arc(x,y+2,30,230,16);
arc(x,y+3,80,240,28);
arc(x+2,y+8,180,330,22);
arc(x-2,y+2,310,460,25);
ellipse(x-12,y+30,120,300,30,40);
ellipse(x+10,y+28,250,423,30,42);
ellipse(x-4,y+10,290,393,30,40);
setfillstyle(SOLID_FILL,4);
floodfill(x+5,y+31,12);
/*画紫色花骨朵*/
ellipse(x+120,y+5,0,360,15,25);
setfillstyle(SOLID_FILL,1);
floodfill(x+120,y,12);
/*画黄色花骨朵*/
ellipse(x-70,y+10,0,360,14,20);
setfillstyle(SOLID_FILL,14);
floodfill(x-70,y+10,12);
setcolor(10);
/*画红花花萼*/
ellipse(x-15,y+32,190,310,30,35);
ellipse(x+16,y+32,235,355,26,35);
ellipse(x,y+35,190,350,43,50);
arc(x,y+82,190,350,6);
setfillstyle(SOLID_FILL,2);
floodfill(x,y+75,10);
/*画粉花花萼*/
ellipse(x+50,y-48,190,320,22,50);
ellipse(x+80,y-48,220,350,22,50);
ellipse(x+65,y-28,180,360,36,50);
floodfill(x+65,y+18,10);
/*画主枝*/
for(i=0;i<3;i++ )
{
ellipse(x-98,y+100+i,255,371,100,80);
ellipse(x-20,y+30+i,260,358,140,140);
ellipse(x+224,y+20+i,180,218,160,140);
}
/*画侧枝*/
ellipse(x+70,y+34,180,233,140,140);
ellipse(x,y+40,205,255,100,120);
ellipse(x+135,y-30,209,249,72,120);
ellipse(x,y+20,263,301,100,120);
ellipse(x+85,y-10,278,305,100,120);
ellipse(x+100,y-62,282,308,90,120);
ellipse(x-50,y-10,277,314,30,120);
ellipse(x+70,y+80,222,266,52,120);
ellipse(x-60,y-45,229,266,52,120);
ellipse(x+79,y-45,229,266,52,120);
ellipse(x+84,y,224,273,52,120);
ellipse(x+110,y+40,240,282,100,120);
/*画紫花骨朵花萼*/
ellipse(x+120,y-6,200,340,17,25);
ellipse(x+120,y+7,160,380,17,27);
floodfill(x+122,y+30,10);
/*画黄花骨朵花萼*/
ellipse(x-70,y+15,140,390,17,20);
ellipse(x-75,y-10,205,340,10,30);
ellipse(x-60,y-10,195,340,5,30);
floodfill(x-70,y+32,10);
/*画叶*/
/*画右边的叶子*/
elli(x+168,y+282,10,20,-40);
elli(x+160,y+252,8,16,260);
elli(x+145,y+270,8,16,-15);
elli(x+156,y+224,10,20,-45);
elli(x+150,y+200,8,16,270);
elli(x+135,y+220,8,16,-10);
elli(x+146,y+144,8,16,-80);
elli(x+130,y+130,6,12,235);
elli(x+125,y+154,7,14,-10);
elli(x+78,y+98,6,12,-90);
elli(x+60,y+90,5,10,180);
elli(x+70,y+109,5,10,-45);
/*画左边的叶子*/
elli(x-125,y+270,12,24,60);
elli(x-95,y+270,10,20,10);
elli(x-110,y+245,10,20,90);
elli(x-105,y+220,10,20,45);
elli(x-100,y+190,8,16,135);
elli(x-75,y+210,8,16,-45);
/*画中间的叶子*/
elli(x+65,y+190,10,20,-45);
elli(x+40,y+185,8,16,0);
elli(x+55,y+165,8,16,90);
}
/*主程序*/
main()
{
int driver=VGA,mode=VGAHI;
registerbgidriver(EGAVGA_driver);
initgraph(&driver,&mode,"C:\TC");
cleardevice();
hua(450,90);
setcolor(2);
rectangle(0,0,639,479);
getch();
closegraph();
}
㈥ 用c语言编辑一个玫瑰花数（一个四位数，各个数字的4次方之和等于它本身，求出满足条件的所有四位数）的
#include <stdio.h>
main()
{
// 从键盘上输入一个四位数，判断其是否为四叶玫瑰数
// （提示：四叶玫瑰数是指一个四位数，它的每个位上的数字的4次幂之和等于它本身）
int rose,one,two,three,four;
scanf("%d",&rose);//
four=rose/1000;
three=rose/100%10;
two=rose/10%10;
one=rose%10;
four=four*four*four*four;
three=three*three*three*three;
two=two*two*two*two;
one=one*one*one*one;
if((four+three+two+one)==rose)
printf("%d",rose);
}
㈦ 输出1000到9999之间的四叶玫瑰数，用C语言的知识回答
可以写成调用函数
#include<stdio.h>
void
rose(int
n)
{int
a,b,c,d;
a=n/1000;
b=n/100%10;
c=n/10%10;
d=n%10;
if(a*a*a*a+b*b*b*b+c*c*c*c+d*d*d*d==n)
printf("%d\t",n);
}
main()
{
int
i;
for(i=1000;i<=9999;i++)
rose(i);
}
方法比较笨，但更容易理解。望楼主采纳。。。