C语言高手帮忙啊!
C\u8bed\u8a00\u9ad8\u624b\u5e2e\u5e2e\u5fd9\u554a\uff01void \u5b66\u751f\u4fe1\u606f\u5904\u7406(\u5b66\u751f\u4fe1\u606f\u7c7b \u73ed\u7ea7\u4fe1\u606f[], int \u4eba\u6570)
{
float \u5e73\u5747\u5206\u6570(0.0);
for (int a = 0; a < \u4eba\u6570; a++)
\u5e73\u5747\u5206\u6570 += \u73ed\u7ea7\u4fe1\u606f[a].\u6210\u7ee9;
cout << \u5e73\u5747\u5206\u6570 / \u4eba\u6570 << endl;
for (int a = 0; a < \u4eba\u6570;a++)
if (\u73ed\u7ea7\u4fe1\u606f[a].\u6210\u7ee9 < 60)
cout << \u73ed\u7ea7\u4fe1\u606f[a].\u59d3\u540d << endl;
};
-----------
cout << "\u8bf7\u8f93\u5165\u7edf\u8ba1\u4eba\u6570:";
int \u6210\u7ee9, \u5b66\u53f7, \u4eba\u6570, \u8ba1\u6570(0); string \u59d3\u540d;
cin >> \u4eba\u6570;
\u5b66\u751f\u4fe1\u606f\u7c7b \u73ed\u7ea7\u4fe1\u606f[100];
cout << "\u8bf7\u8f93\u5165\u5b66\u751f\u4fe1\u606f:" << endl;
do
{
cin >> \u5b66\u53f7 >> \u59d3\u540d >> \u6210\u7ee9;
\u73ed\u7ea7\u4fe1\u606f[\u8ba1\u6570] = { \u5b66\u53f7, \u59d3\u540d, \u6210\u7ee9 };
\u8ba1\u6570++;
} while (\u4eba\u6570 > \u8ba1\u6570);
\u5b66\u751f\u4fe1\u606f\u5904\u7406(\u73ed\u7ea7\u4fe1\u606f, \u4eba\u6570);
#include
void main(){
void eliminate(int n, int k);
int total, num;
printf("\u8bf7\u8f93\u5165\u603b\u4eba\u6570\uff1a\n");
scanf("%d",&total);
printf("\u8bf7\u8f93\u5165\u6dd8\u6c70\u53f7k\uff1a\n");
scanf("%d",&num);
eliminate(total,num);
}
void eliminate(int n, int k){
int *arr, size=n, count=-1;
arr=new int[n];
for(int i=0; i<n; i++){
arr[i]=i+1;//\u8fd9\u91cc\u53ef\u4ee5\u7528\u8f93\u5165\u8bed\u53e5scanf("%d",&a[i]);\u6765\u624b\u52a8\u8f93\u5165\u6bcf\u4e2a\u53c2\u4e0e\u8005\u7684\u7f16\u53f7
}
while(size>1){
for(int i=0;i<k;i++){
count++;
if(count==size)
count=0;
}
size--;
printf("\u7b2c %d \u4e2a\u6dd8\u6c70\u8005\u662f %d \u53f7\n",n-size,arr[count]);
if(count!=size)
for(int i=count;i<=size;i++)
arr[i]=arr[i+1];
else
count=0;
}
printf("\u5e78\u5b58\u8005\u662f %d \u53f7\n",arr[0]);
}
#include <stdio.h>
#include <math.h>
struct point
{
float x;
float y;
};
get_point(struct point *p)
{
float a,b;
scanf("%f,%f",&a,&b);
p->x=a;
p->y=b;
return 0;
}
put_point(struct point a)
{
printf("%f,%f\n",a.x,a.y);
return 0;
}
struct point add(struct point a,struct point b)
{
struct point c;
c.x=a.x+b.x;
c.y=a.y+b.y;
return c;
}
struct point sub(struct point a,struct point b)
{
struct point c;
c.x=a.x-b.x;
c.y=a.y-b.y;
return c;
}
float triangle(struct point A, struct point B, struct point C)
{
float a,b,c,p,area;
struct point d,e,f;
d=sub(A,B);
e=sub(A,C);
f=sub(B,C);
c=sqrt(d.x*d.x+d.y*d.y);
b=sqrt(e.x*e.x+e.y*e.y);
a=sqrt(f.x*f.x+f.y*f.y);
p=(a+b+c)/2;
area=sqrt(p*(p-a)*(p-b)*(p-c));
return area;
}
main()
{
struct point a,b,c,n;
printf("Please input a point:");
get_point(&a);
printf("The Point you input is: ");
put_point(a);
printf("Please input 3 points of triangle :");
get_point(&a);
get_point(&b);
get_point(&c);
printf("The area of triangle is:%f\n",triangle(a,b,c));
printf("输入平移向量坐标:");
get_point(&n);
a=add(a,n);
b=add(b,n);
c=add(c,n);
printf("平移后三角形面积为:%f",triangle(a,b,c));
}
/*旋转*/
#include
#include
#define DIM 4
#define N 1
#define NUM 3
#define PI 3.1415926
typedef struct POINT
{
int x,y,z;
}POINT;
void trans(double matrix[DIM][DIM], double x, double y, double z) //矩阵平移
{
for(int i=0;i {
for(int j=0;j {
if(i==j) matrix[i][j]=1;
else matrix[i][j]=0;
}
}
matrix[0][3]=x;
matrix[1][3]=y;
matrix[2][3]=z;
return;
}
void t(double matrix[][DIM],double matrix1[][DIM]) //矩阵的转置
{
for(int i=0;i {
for(int j=0;j {
matrix[i][j]=matrix1[j][i];
}
}
return;
}
void rotate(double matrix[][DIM],int angle) //矩阵绕 z 坐标轴旋转
{
int n=360;
for(int i=0;i {
for(int j=0;j {
if(i==j) matrix[i][j]=1;
else matrix[i][j]=0;
}
}
matrix[0][0]=cos(2*PI/n*angle);
matrix[1][1]=matrix[0][0];
matrix[0][1]=-sin(2*PI/n*angle);
matrix[1][0]=-matrix[0][1];
return;
}
void new_or(double matrix[][DIM],int px1,int py1,int pz1,int px2,int py2,int pz2) //根据旋转向量,建立坐标系
{
for(int i=0;i {
for(int j=0;j {
if(i==j) matrix[i][j]=1;
else matrix[i][j]=0;
}
}
double px,py,pz,mo,mo1;
px=px2-px1;
py=py2-py1;
pz=pz2-pz1;
mo=sqrt(px*px+py*py+pz*pz);
matrix[2][0]=px/mo;
matrix[2][1]=py/mo;
matrix[2][2]=pz/mo;
mo1=sqrt(px*px+py*py);
matrix[0][0]=-py/mo1;
matrix[0][1]=px/mo1;
matrix[0][2]=0;
matrix[1][0]=matrix[2][1]*matrix[0][2]-matrix[2][2]*matrix[0][1];
matrix[1][1]=matrix[2][2]*matrix[0][0]-matrix[0][2]*matrix[2][0];
matrix[1][2]=matrix[2][0]*matrix[0][1]-matrix[0][0]*matrix[2][1];
return;
}
void phalanx_mul(double matrix[][DIM],double matrix1[][DIM],double matrix2[][DIM])//方阵乘法4*4 * 4*4
{
double mul[DIM][DIM];
for(int i=0;i {
for(int j=0;j {
mul[i][j]=0;
for(int k=0;k {
mul[i][j]+=matrix1[i][k]*matrix2[k][j];
}
}
}
for(i=0;i {
for(int j=0;j {
matrix[i][j]=mul[i][j];
}
}
}
void matrix_mul(double matrix[DIM][N],double matrix1[DIM][DIM],double matrix2[DIM][N])//矩阵乘法4*4 * 4*1
{
for(int i=0;i {
for(int j=0;j {
matrix[i][j]=0;
for(int k=0;k {
matrix[i][j]+=matrix1[i][k]*matrix2[k][j];
}
}
}
return;
}
void main()
{
POINT startp,endp,pt[NUM];
int angle;
double trans1[DIM][DIM]={{1,0,0,0},{0,1,0,0},{0,0,1,0},{0,0,0,1}},
trans2[DIM][DIM]={{1,0,0,0},{0,1,0,0},{0,0,1,0},{0,0,0,1}},
matrix[DIM][DIM]={{1,0,0,0},{0,1,0,0},{0,0,1,0},{0,0,0,1}},
rotate1[DIM][DIM]={{1,0,0,0},{0,1,0,0},{0,0,1,0},{0,0,0,1}},
matrix1[DIM][N]={0},point[DIM][N]={0},
t1[DIM][DIM]={{1,0,0,0},{0,1,0,0},{0,0,1,0},{0,0,0,1}},
result[DIM][DIM]={{1,0,0,0},{0,1,0,0},{0,0,1,0},{0,0,0,1}};
printf("请输入 %d 个点代表三角形的三个顶点:\n",NUM);//构建三角形的三个顶点
for(int i=0;i {
printf("pt[%d].x:\t",i);
scanf("%d",&pt[i].x );
printf("pt[%d].y:\t",i);
scanf("%d",&pt[i].y );
printf("pt[%d].z:\t",i);
scanf("%d",&pt[i].z );
}
printf("输入旋转轴向量的起始坐标和终止坐标:\n"); //构建旋转轴坐标
printf("startp.x:\t");
scanf("%d",&startp.x );
printf("startp.y:\t");
scanf("%d",&startp.y );
printf("startp.z:\t");
scanf("%d",&startp.z );
printf("endp.x:\t\t");
scanf("%d",&endp.x );
printf("endp.y:\t\t");
scanf("%d",&endp.y );
printf("endp.z:\t\t");
scanf("%d",&endp.z );
printf("\n输入旋转的角度:\n"); //旋转角度
scanf("%d",&angle);
if(startp.x !=0 && startp.y !=0 && startp.z !=0) //假如向量起始坐标非零
{
trans(trans1, -startp.x, -startp.y, -startp.z);
trans(trans2, startp.x, startp.y, startp.z);
}
printf("\n平移矩阵的逆\n");
for(i=0;i {
for(int j=0;j {
printf("%f\t",trans1[i][j]);
}
printf("\n");
}
printf("\n平移矩阵\n");
for(i=0;i {
for(int j=0;j {
printf("%f\t",trans2[i][j]);
}
printf("\n");
}
new_or(matrix,startp.x ,startp.y ,startp.z ,endp.x ,endp.y ,endp.z );
printf("\n构建新坐标系A\n");
for(i=0;i {
for(int j=0;j {
printf("%f\t",matrix[i][j]);
}
printf("\n");
}
t(t1,matrix);
printf("\n求新坐标系A的转置（逆）\n");
for(i=0;i {
for(int j=0;j {
printf("%f\t",t1[i][j]);
}
printf("\n");
}
rotate(rotate1,angle);
printf("\n旋转矩阵\n");
for(i=0;i {
for(int j=0;j {
printf("%f\t",rotate1[i][j]);
}
printf("\n");
}
phalanx_mul(result,trans2,t1);
printf("\n平移的逆 * A的转置\n");
for(i=0;i {
for(int j=0;j {
printf("%f\t",result[i][j]);
}
printf("\n");
}
phalanx_mul(result,result,rotate1);
printf("\n平移的逆 * A的转置 * 旋转矩阵R\n");
for(i=0;i {
for(int j=0;j {
printf("%f\t",result[i][j]);
}
printf("\n");
}
phalanx_mul(result,result,matrix);
printf("\n平移的逆 * A的转置 * 旋转矩阵R * A\n");
for(i=0;i {
for(int j=0;j {
printf("%f\t",result[i][j]);
}
printf("\n");
}
phalanx_mul(result,result,trans1);
printf("\n平移的逆 * A的转置 * 旋转矩阵R * A * 平移\n");
for(i=0;i {
for(int j=0;j {
printf("%f\t",result[i][j]);
}
printf("\n");
}
for(int k=0;k {
printf("\n旋转后点%d的坐标为:\n",k);
matrix1[0][0]=pt[k].x;
matrix1[1][0]=pt[k].y;
matrix1[2][0]=pt[k].z;
matrix1[3][0]=1;
matrix_mul(point,result,matrix1);
for(i=0;i {
for(int j=0;j {
printf("%f\t",point[i][j]);
}
printf("\n");
}
}
}
这么复杂也不给点分,没有人愿意回答的.
就是，太抠了。
【1】 矩阵的运算简单，按照定义，熟悉语言就可以做，又不需要优化算法。
【2】平移：三角形的三个顶点的横纵坐标分别加上对应的平移值。伸缩：如果以其中一个顶点的位置坐标不变的话，该顶点的两条边伸缩求另外两个顶点。旋转：安旋转角度求坐标，要看旋转的参照点。
C语言仅仅是工具！
【1】学过还好做，【2】不会；
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