c语言给定半径,求其对应的圆周长.面积,对应球体体积:使用宏定义3.1415

voidmain(){floatr,s,l;floatpi=3.1415;printf("输入一个圆的半径:\n");scanf("%f",&r);if(

#includevoidmain(){\x05intn;\x05intsum=0;\x05scanf("%d",&n);\x05while(n)\x05{\x05\x05sum+=n%10;\x05\

voidmain(){inta[10]={1,2,3,4,5,6,7,8,9,10};inta=0,b=0,i=0;for(i=0;i

#include#defineRow50//行数最大值voidmain(){inti=0;intj=0;//j为临时变量intarray[Row][101];//二维数组//输入数组do{printf

#includevoidmain(){inta,b,c;intn;doubleimport;while(scanf("%d",&n)!=EOF&&n!=0){//instala=b=c=0;while

#includeintmain(){\x09inta=0,b=0,c=0,n,m;//分别表示正数,负数,零\x09printf("请输入数字的个数\n");\x09scanf("%d",&n);\x

#includemain(){floatpi,r,h,l,s,sq,vq,vz;pi=3.141592654;printf("r=,h=\n");scanf("%f,%f",&r,&h);l=2*pi

由圆周率的含义可知：圆周率π表示圆的周长和它直径的比值；故选：A．

#include#include#include#include#defineN9//N个节点#defineM2//M次拟合#defineK2*Mvoidzhuyuan(intk,intn,float

void main(){int n1,n2;printf("输入第1个整数:n1\n");scanf("%d",&n1);p

#...intr,b,dsin

一个数n若(n&(n-1))==0则n是2的幂再问：为什么啊

改过的,有注释,对照看……voidmain(void){longinta[100000],x,y,i,j,k,z,m,s;scanf("%d%d",&x,&y);z=y-x-1;m=0;for(i=0

#includeintScanDMatrix(doublematrix[][2]);voidPrintDMatrix(doublematrix[][2]);intInverseDMatrix(doub

平面是怎么个表达法图像么是以二维矩阵存储么再问：这个不知道啊你就看着题目办吧再答：那两个点有x,y坐标吧,point1，point2分别为点1和点2传入4个坐标值返回距离注意要有#includedou

#include#include#includeusingnamespacestd;intmain(){intr=15,h=3;doublec=2*3.14*r;doubles=pow(r,2)*3.

设角度为θ,则坐标就是（rcos+x,rsinθ+y)

printf("area=3.14*r*r;cir=2*3.14*r;\n);你这样写肯定错了,printf的输出格式错了.应该这样写printf("%f,%f\n",area=3.14*r*r,ci

#includeintmain(){\x09inta,b,i;\x09scanf("%d%d",&a,&b);\x09while(0

#include#include#include#includeusingnamespacestd;doubledet(intN,vectorA){doubleD=0;vectorB((N-1)*(N