c语言a算法
1、有穷性（有限性）。任何一种提出的解题方法都是在有限的操作步骤内可以完成的。
如果在有限的操作步骤内完不成，得不到结果，这样的算法将无限的执行下去，永远不会停止。除非手动停止。例如操作系统就不具有有穷性，它可以一直运行。
2、一个算法应该具有以下七个重要的特征：
1）有穷性（finiteness）
算法的有穷性是指算法必须能在执行有限个步骤之后终止
2）确切性(definiteness)
算法的每一步骤必须有确切的定义；
3）输入项(input)
一个算法有0个或多个输入,以刻画运算对象的初始情况,所谓0个输入是指算法本身定出了初始条件；
4）输出项(output)
一个算法有一个或多个输出,以反映对输入数据加工后的结果.没有输出的算法是毫无意义的；
5）可行性(effectiveness)
算法中执行的任何计算步都是可以被分解为基本的可执行的操作步,即每个计算步都可以在有限时间内完成；
6）
高效性（high
efficiency）
执行速度快,占用资源少；
7）
健壮性（robustness）
健壮性又称鲁棒性，是指软件对于规范要求以外的输入情况的处理能力。所谓健壮的系统是指对于规范要求以外的输入能够判断出这个输入不符合规范要求，并能有合理的处理方式。
⑵ c语言的ASCII怎么算法
c里头，char和int同属于“整型”,他们都可以用整数来表示。char的整数表示就是ascii里头对应的那个整数，所以，char也可以进行加减运算。
ascii表中，小写字母在大写字母后面，而且中间间隔有其他符号，'a'-'A'表示的就是对应两个大写字母和小写字母之间的差值。那么c2是大写的A也就不难理解了。
⑶ 求A* 算法C语言源程序
#include <string.h>
#include <stdio.h>
#include <malloc.h>
#include <time.h>
#define NULL 0
const int nmax = 200;
const int nend = 99; /*终点坐标的代表点*/
static char achar[10][10];
static int npayo = 0; /*0 表示空 1为非空*/
static int npayc = 0; /*0 表示空 1为非空*/
static int npay_x = 0; /*起点*/
static int npay_y = 0;
static int nend_x = 9; /*终点*/
static int nend_y = 9;
static int nnewpay_x;
static int nnewpay_y;
static int ndian = 101;
static int nh;
static long i = 10000000L;
struct Spaydian
{
int ng;
int nf; /*路径评分*/
int nmy_x; /*自己位置*/
int nmy_y;
int nfatherx; /*父节点*/
int nfathery;
int nflag; /* 0 wei O; 1 wei @ */
};
static struct Spaydian spaydian[200];
/* open close list 表 */
typedef struct spaylist
{
int n_f;
int n_x;
int n_y;
int nfather_x;
int nfather_y;
struct spaylist *next;
};
static struct spaylist *open_list, *close_list;
static void smline();
static int sjudge(int nx,int ny,int i); /*判断在第nx列ny行向第i个方向走是否可以,可以返回1否则返回0。
i=1表示向右，2表示向下，3表示向左，4表示向上*/
static void spath();
static void spayshow(int nxx,int nyy);
static int sifopen( int nx,int ny); /*判断点是否在 open 表上*/
static int sifclose(int nx,int ny); /*判断点是否在 close 表上*/
static int snewx(int nx,int i);
static int snewy(int ny,int i);
static spaylist *creat(); /*建立链表*/
static spaylist *del(spaylist *head,int num_x,int num_y); /*删除链表的结点*/
static spaylist *snewdianx(spaylist *head);/*新的点*/
static spaylist *snewdiany(spaylist *head);
static spaylist *insert(spaylist *head,int ndian); /* 点插入链表 */
static spaylist *srebirth(spaylist *head,int ndian); /*更新表*/
int main()
{
FILE *fp ;
char ach ;
int ni = 0 ; /*统计个数*/
int nii = 0; /*achar[nii][njj]*/
int njj = 0;
if ((fp=fopen("route.txt","rt")) == NULL) /* 判断打开文件是否为空 */
{
printf("文件为空！~\n");
return 0;
/* exit(1);*/
}
ach = fgetc(fp);
while(ach != EOF)
{
if(ach == 'O' || ach == '@') /*当值为@或O时候*/
{
spaydian[ni].ng = 0;
spaydian[ni].nf = nmax;
spaydian[ni].nmy_x = njj;
spaydian[ni].nmy_y = nii;
spaydian[ni].nfathery = -1;
spaydian[ni].nfatherx = -1;
if(ach == '@')
{
spaydian[ni].nflag = 1;
}
else if(ach == 'O')
{
spaydian[ni].nflag = 0;
}
ni++;
achar[nii][njj] = ach;
njj++;
if(njj == 10)
{
nii++;
njj = 0;
}
} /*end if*/
ach = fgetc(fp);
}/*end while*/
smline(); /* a*算法 */
fp=fopen("answer.txt","w");
for(int i=0;i<10;i++ )
{ for(int j=0;j<10;j++ )
{
printf("%c",achar[i][j]);
if(j==9)
printf("\n");
fprintf(fp,"%c",achar[i][j]);
if (j==9)
fprintf(fp,"\n");
}
}
fclose(fp);
return 0;
}
/* a* 算法 */
static void smline()
{ close_list = open_list = NULL;
open_list = creat();
while(open_list != NULL) /* 当open 表不为空时 */
{
open_list = del(open_list,npay_x,npay_y); /*删除open 链表的结点*/
if(npay_x == 9 && npay_y == 9)
{
achar[9][9] = '=';
spath(); /*寻找并画出路径*/
break;
}
for (int i=1; i<=4; i++) /*四个方向逐个行走，i=1向右 2向下 3向左 4向上*/
{
if (sjudge(npay_x,npay_y,i))
{
nnewpay_x = snewx(npay_x,i);
nnewpay_y = snewy(npay_y,i);
if(open_list != NULL)
npayo = sifopen(nnewpay_x,nnewpay_y) ; /*判断点是否在 open 表中*/
else
npayo = 0;
if(close_list != NULL)
npayc = sifclose(nnewpay_x,nnewpay_y) ; /*判断点是否在 close 表中*/
else
npayc = 0;
ndian = 10*nnewpay_x+nnewpay_y ;
if (npayo == 0 && npayc == 0 ) /*点不在open表也不在close表中*/
{
spaydian[ndian].ng = spaydian[10*npay_x+npay_y].ng + 1; /*更新点的基本信息*/
nh = (nend - ndian)/10 + (nend - ndian)%10 ;
spaydian[ndian].nf = spaydian[ndian].ng+nh;
spaydian[ndian].nfathery = npay_y;
spaydian[ndian].nfatherx = npay_x;
spaydian[ndian].nmy_y = nnewpay_y;
spaydian[ndian].nmy_x = nnewpay_x;
open_list = insert(open_list,ndian);/*点插入open 表中*/
}
else if (npayo == 1) /*点在open表中*/
{
spaydian[ndian].ng = spaydian[10*npay_x+npay_y].ng + 1;
nh = (nend - ndian)/10 + (nend - ndian)%10 ;
if(spaydian[ndian].nf > (spaydian[ndian].ng+nh) && spaydian[ndian].nf != nmax)
{
spaydian[ndian].nf = spaydian[ndian].ng+nh;
open_list = srebirth(open_list,ndian); /*点插入open 表中*/
}
}
else if(npayc == 1) /*新生成的点在close表中*/
{
spaydian[ndian].ng = spaydian[10*npay_x+npay_y].ng + 1;
nh = (nend - ndian)/10 + (nend - ndian)%10 ;
if(spaydian[ndian].nf > (spaydian[ndian].ng+nh) && spaydian[ndian].nf != nmax)
{
spaydian[ndian].nf = spaydian[ndian].ng+nh;
close_list = srebirth(close_list,ndian);
close_list = del(close_list,nnewpay_x,nnewpay_y); /*删除close链表的结点*/
open_list = insert(open_list,ndian);/*点插入open 表中*/
}
}/*end else if*/
}/*end if*/
}/*end for*/
close_list = insert(close_list,(10*npay_x+npay_y));/*点插入close 表中*/
if(open_list != NULL)
{
npay_x = open_list->n_x;
npay_y = open_list->n_y;
}
}/*end while*/
if(open_list == NULL)
{printf("无路可走 \n");}
}
/*建立链表*/
spaylist *creat(void)
{
spaylist *head;
spaylist *p1;
int n=0;
p1=(spaylist*)malloc(sizeof(spaylist));
p1->n_f = 18;
p1->n_x = 0;
p1->n_y = 0;
p1->nfather_x = -1;
p1->nfather_x = -1;
p1->next = NULL;
head = NULL;
head=p1;
return(head);
}
/*删除结点*/
spaylist *del(spaylist *head,int num_x,int num_y)
{
spaylist *p1, *p2;
if(head == NULL)
{
printf("\nlist null!\n");
return (head);
}
p1 = head;
while((num_y != p1->n_y ||num_x != p1->n_x )&& p1->next != NULL)
{
p2=p1;
p1=p1->next;
}
if(num_x == p1->n_x && num_y == p1->n_y )
{
if(p1==head)
head=p1->next;
else
p2->next=p1->next;
}
return (head);
}
/*输出*/
static void spath()
{
int nxx;
int nyy;
nxx = spaydian[nend].nfatherx;
nyy = spaydian[nend].nfathery;
spayshow(nxx,nyy) ;
}
/*递归*/
void spayshow(int nxx,int nyy)
{ achar[nxx][nyy] = '=';
if( nxx != 0 || nyy != 0 )
{
int nxxyy = 10*nxx+nyy;
nxx = spaydian[nxxyy].nfatherx;
nyy = spaydian[nxxyy].nfathery;
spayshow(nxx,nyy);
}
}
/* 判断周围四个点是否可行 */
static int sjudge(int nx,int ny,int i)
{
if (i==1) /*判断向右可否行走*/
{
if (achar[nx][ny+1]=='O' && ny<9)
{
return 1;
}
else
return 0;
}
else if (i==2) /*判断向下可否行走*/
{
if (achar[nx+1][ny]=='O' && nx<9)
{
return 1;
}
else
return 0;
}
else if (i==3)/*判断向左可否行走 */
{
if (ny > 0&&achar[nx][ny-1]=='O')
{
return 1;
}
else
return 0;
}
else if (i==4)/*判断向上可否行走 */
{
if (nx > 0&&achar[nx-1][ny]=='O')
{
return 1;
}
else
return 0;
}
else
return 0;
}
/* 求新的x点 */
static int snewx(int nx,int i)
{
if(i == 1)
nx = nx;
else if(i == 2)
nx = nx+1;
else if(i == 3)
nx = nx;
else if(i == 4)
nx = nx-1;
return nx;
}
/* 求新的y点 */
static int snewy(int ny, int i)
{
if(i == 1)
ny = ny+1;
else if(i == 2)
ny = ny;
else if(i == 3)
ny = ny-1;
else if(i == 4)
ny = ny;
return ny;
}
/*判定点是否在open表中*/
int sifopen(int nx,int ny)
{
spaylist *p1, *p2;
p1 = open_list;
while((ny != p1->n_y || nx != p1->n_x )&& p1->next != NULL)
{
p2 = p1;
p1 = p1->next;
}
if(nx == p1->n_x && ny == p1->n_y)
return 1;
else
return 0;
}
/*判定点是否在close表中*/
int sifclose(int nx,int ny)
{
spaylist *p1, *p2;
p1 = close_list;
while((ny != p1->n_y ||nx != p1->n_x )&& p1->next != NULL)
{
p2=p1;
p1=p1->next;
}
if(nx == p1->n_x && ny == p1->n_y)
return 1;
else
return 0;
}
/*插入结点*/
spaylist * insert(spaylist *head,int ndian)
{
spaylist *p0,*p1,*p2;
p1=head;
p0=(spaylist*)malloc(sizeof(spaylist));
p0->n_f = spaydian[ndian].nf;
p0->n_x = spaydian[ndian].nmy_x;
p0->n_y = spaydian[ndian].nmy_y;
p0->nfather_x = spaydian[ndian].nfatherx;
p0->nfather_x = spaydian[ndian].nfathery;
p0->next = NULL;
if(head==NULL)
{
head=p0;
p0->next=NULL;
}
else
{
while((p0->n_f > p1->n_f)&&(p1->next!=NULL))
{
p2=p1;
p1=p1->next;
}
if(p0->n_f <= p1->n_f)
{
if(head==p1)
head=p0;
else
p2->next=p0;
p0->next=p1;
}
else
{
p1->next=p0;
p0->next=NULL;
}
}
return (head);
}
/* 更新链表 */
spaylist * srebirth(spaylist *head,int ndian)
{
spaylist *p1, *p2;
p1=head;
while(spaydian[ndian].nmy_x!=p1->n_x&&spaydian[ndian].nmy_x!=p1->n_x&&p1->next!=NULL)
{
p2=p1;
p1=p1->next;
}
if(spaydian[ndian].nmy_x==p1->n_x&&spaydian[ndian].nmy_x==p1->n_x)
{
p1->n_f = spaydian[ndian].nf;
}
return (head);
}
⑷ 用C语言怎么编一个a^n(a的n次方)的算法
如果n比较小，可以吧
result
*=
a循环n次。。
如果n比较大，
可以逐步来算。
这样考虑，f(n)
=
2^n
如果有了
f(m)的结果，
那么
f(2m)和f(2m+1)
就分别等于
f(m)*f(m)和f(m)*f(m)*a
所以可以从最高位开始查看n的每一位，
如果这一位是1，
那么
result
=
result
*
result
*
a;
如果这一位是0，那么result
=
result
*
result;
其中result
的初始值是1。
这样复杂度就是log(n)的
⑸ C语言a=a++的运算顺序是怎么样的
这两个程序的输出结果是相同的：
因为它们的操作都是：先取变量a的值，取完后a自增，最后取前面取到的值赋值给赋值号左边的变量（所以最后输出变量的值就都是1）。
⑹ A*搜寻算法的代码实现(C语言实现)
用C语言实现A*最短路径搜索算法，作者 Tittup frog(跳跳蛙)。 #include<stdio.h>#include<math.h>#defineMaxLength100 //用于优先队列（Open表）的数组#defineHeight15 //地图高度#defineWidth20 //地图宽度#defineReachable0 //可以到达的结点#defineBar1 //障碍物#definePass2 //需要走的步数#defineSource3 //起点#defineDestination4 //终点#defineSequential0 //顺序遍历#defineNoSolution2 //无解决方案#defineInfinity0xfffffff#defineEast(1<<0)#defineSouth_East(1<<1)#defineSouth(1<<2)#defineSouth_West(1<<3)#defineWest(1<<4)#defineNorth_West(1<<5)#defineNorth(1<<6)#defineNorth_East(1<<7)typedefstruct{ signedcharx,y;}Point;constPointdir[8]={ {0,1},//East {1,1},//South_East {1,0},//South {1,-1},//South_West {0,-1},//West {-1,-1},//North_West {-1,0},//North {-1,1}//North_East};unsignedcharwithin(intx,inty){ return(x>=0&&y>=0 &&x<Height&&y<Width);}typedefstruct{ intx,y; unsignedcharreachable,sur,value;}MapNode;typedefstructClose{ MapNode*cur; charvis; structClose*from; floatF,G; intH;}Close;typedefstruct//优先队列（Open表）{ intlength; //当前队列的长度 Close*Array[MaxLength]; //评价结点的指针}Open;staticMapNodegraph[Height][Width];staticintsrcX,srcY,dstX,dstY; //起始点、终点staticCloseclose[Height][Width];//优先队列基本操作voidinitOpen(Open*q) //优先队列初始化{ q->length=0; //队内元素数初始为0}voidpush(Open*q,Closecls[Height][Width],intx,inty,floatg){ //向优先队列（Open表）中添加元素 Close*t; inti,mintag; cls[x][y].G=g; //所添加节点的坐标 cls[x][y].F=cls[x][y].G+cls[x][y].H; q->Array[q->length++]=&(cls[x][y]); mintag=q->length-1; for(i=0;i<q->length-1;i++) { if(q->Array[i]->F<q->Array[mintag]->F) { mintag=i; } } t=q->Array[q->length-1]; q->Array[q->length-1]=q->Array[mintag]; q->Array[mintag]=t; //将评价函数值最小节点置于队头}Close*shift(Open*q){ returnq->Array[--q->length];}//地图初始化操作voidinitClose(Closecls[Height][Width],intsx,intsy,intdx,intdy){ //地图Close表初始化配置 inti,j; for(i=0;i<Height;i++) { for(j=0;j<Width;j++) { cls[i][j].cur=&graph[i][j]; //Close表所指节点 cls[i][j].vis=!graph[i][j].reachable; //是否被访问 cls[i][j].from=NULL; //所来节点 cls[i][j].G=cls[i][j].F=0; cls[i][j].H=abs(dx-i)+abs(dy-j); //评价函数值 } } cls[sx][sy].F=cls[sx][sy].H; //起始点评价初始值 // cls[sy][sy].G=0; //移步花费代价值 cls[dx][dy].G=Infinity;}voidinitGraph(constintmap[Height][Width],intsx,intsy,intdx,intdy){ //地图发生变化时重新构造地 inti,j; srcX=sx; //起点X坐标 srcY=sy; //起点Y坐标 dstX=dx; //终点X坐标 dstY=dy; //终点Y坐标 for(i=0;i<Height;i++) { for(j=0;j<Width;j++) { graph[i][j].x=i;//地图坐标X graph[i][j].y=j;//地图坐标Y graph[i][j].value=map[i][j]; graph[i][j].reachable=(graph[i][j].value==Reachable); //节点可到达性 graph[i][j].sur=0;//邻接节点个数 if(!graph[i][j].reachable) { continue; } if(j>0) { if(graph[i][j-1].reachable) //left节点可以到达 { graph[i][j].sur|=West; graph[i][j-1].sur|=East; } if(i>0) { if(graph[i-1][j-1].reachable &&graph[i-1][j].reachable &&graph[i][j-1].reachable) //up-left节点可以到达 { graph[i][j].sur|=North_West; graph[i-1][j-1].sur|=South_East; } } } if(i>0) { if(graph[i-1][j].reachable) //up节点可以到达 { graph[i][j].sur|=North; graph[i-1][j].sur|=South; } if(j<Width-1) { if(graph[i-1][j+1].reachable &&graph[i-1][j].reachable &&map[i][j+1]==Reachable)//up-right节点可以到达 { graph[i][j].sur|=North_East; graph[i-1][j+1].sur|=South_West; } } } } }}intbfs(){ inttimes=0; inti,curX,curY,surX,surY; unsignedcharf=0,r=1; Close*p; Close*q[MaxLength]={&close[srcX][srcY]}; initClose(close,srcX,srcY,dstX,dstY); close[srcX][srcY].vis=1; while(r!=f) { p=q[f]; f=(f+1)%MaxLength; curX=p->cur->x; curY=p->cur->y; for(i=0;i<8;i++) { if(!(p->cur->sur&(1<<i))) { continue; } surX=curX+dir[i].x; surY=curY+dir[i].y; if(!close[surX][surY].vis) { close[surX][surY].from=p; close[surX][surY].vis=1; close[surX][surY].G=p->G+1; q[r]=&close[surX][surY]; r=(r+1)%MaxLength; } } times++; } returntimes;}intastar(){ //A*算法遍历 //inttimes=0; inti,curX,curY,surX,surY; floatsurG; Openq;//Open表 Close*p; initOpen(&q); initClose(close,srcX,srcY,dstX,dstY); close[srcX][srcY].vis=1; push(&q,close,srcX,srcY,0); while(q.length) { //times++; p=shift(&q); curX=p->cur->x; curY=p->cur->y; if(!p->H) { returnSequential; } for(i=0;i<8;i++) { if(!(p->cur->sur&(1<<i))) { continue; } surX=curX+dir[i].x; surY=curY+dir[i].y; if(!close[surX][surY].vis) { close[surX][surY].vis=1; close[surX][surY].from=p; surG=p->G+sqrt((curX-surX)*(curX-surX)+(curY-surY)*(curY-surY)); push(&q,close,surX,surY,surG); } } } //printf("times:%d ",times); returnNoSolution;//无结果}constintmap[Height][Width]={ {0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,1,1}, {0,0,1,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1}, {0,0,0,0,0,0,1,0,0,0,0,0,0,1,1,0,0,0,0,1}, {0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,1}, {0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,1,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0}, {0,1,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0}, {0,0,0,0,1,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0}, {0,1,0,0,0,0,1,0,0,0,0,0,0,1,0,1,0,0,0,1}, {0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0}};constcharSymbol[5][3]={"□","▓","▽","☆","◎"};voidprintMap(){ inti,j; for(i=0;i<Height;i++) { for(j=0;j<Width;j++) { printf("%s",Symbol[graph[i][j].value]); } puts(""); } puts("");}Close*getShortest(){ //获取最短路径 intresult=astar(); Close*p,*t,*q=NULL; switch(result) { caseSequential: //顺序最近 p=&(close[dstX][dstY]); while(p) //转置路径 { t=p->from; p->from=q; q=p; p=t; } close[srcX][srcY].from=q->from; return&(close[srcX][srcY]); caseNoSolution: returnNULL; } returnNULL;}staticClose*start;staticintshortestep;intprintShortest(){ Close*p; intstep=0; p=getShortest(); start=p; if(!p) { return0; } else { while(p->from) { graph[p->cur->x][p->cur->y].value=Pass; printf("（%d，%d）→ ",p->cur->x,p->cur->y); p=p->from; step++; } printf("（%d，%d） ",p->cur->x,p->cur->y); graph[srcX][srcY].value=Source; graph[dstX][dstY].value=Destination; returnstep; }}voidclearMap(){ //ClearMapMarksofSteps Close*p=start; while(p) { graph[p->cur->x][p->cur->y].value=Reachable; p=p->from; } graph[srcX][srcY].value=map[srcX][srcY]; graph[dstX][dstY].value=map[dstX][dstY];}voidprintDepth(){ inti,j; for(i=0;i<Height;i++) { for(j=0;j<Width;j++) { if(map[i][j]) { printf("%s",Symbol[graph[i][j].value]); } else { printf("%2.0lf",close[i][j].G); } } puts(""); } puts("");}voidprintSur(){ inti,j; for(i=0;i<Height;i++) { for(j=0;j<Width;j++) { printf("%02x",graph[i][j].sur); } puts(""); } puts("");}voidprintH(){ inti,j; for(i=0;i<Height;i++) { for(j=0;j<Width;j++) { printf("%02d",close[i][j].H); } puts(""); } puts("");}intmain(intargc,constchar**argv){ initGraph(map,0,0,0,0); printMap(); while(scanf("%d%d%d%d",&srcX,&srcY,&dstX,&dstY)!=EOF) { if(within(srcX,srcY)&&within(dstX,dstY)) { if(shortestep=printShortest()) { printf("从（%d，%d）到（%d，%d）的最短步数是:%d ", srcX,srcY,dstX,dstY,shortestep); printMap(); clearMap(); bfs(); //printDepth(); puts((shortestep==close[dstX][dstY].G)?"正确":"错误"); clearMap(); } else { printf("从（%d，%d）不可到达（%d，%d） ", srcX,srcY,dstX,dstY); } } else { puts("输入错误！"); } } return(0);}
⑺ C语言算法a(n)=2^n+1过程怎么写！急！
以下程序以n>=0的整数为标准。
#include <stdio.h>
int main(){
int a=1,i=0;
int n;
printf("请输入n的值");
scnaf("%d",&n);
for (i=0;i<n;i++)
{a=a*2;}
printf("a=%d",a);
return 0;}
⑻ C语言基本算法
1.输入语句:scanf("控制格式",接受值列表),其中控制格式常用的有:%d,%c,%s,%f,分别
表示整型,字符型,字符串和浮点型.
例如int
a;char
c;scanf("%d
%c",&a,&c);表示向a和c输入值
2.赋值语句:=号,如将b赋值为10,为b=10
3.条件:if(布尔表达式){程序}else{程序}(注:此结构可嵌套)
switch(离散量){case
常量:...;case
常量:...}
例:int
a;scanf("%d",&a);
if(a>10)
{printf("大于10");}
else
{printf("小于10")}
例:switch(months)
{
case
1:printf("1月有31天");break;
case
3:printf("3月有31天");break;
....
default:break;
}
4.循环:for结构,while结构,do-while结构
for(初始化;判断;变化)
{
}
while(条件)
{
}
do
{
}while(条件)
⑼ c语言中a+=a-=a*=a这个表达式的算法是怎么算的
a的初值呢？
a初值为12时，a+=a-=a*=a结果为0
步骤：
这个表达式的运算是从右向左的：
1.
a*=a：a=a*a=12*12=144
2.
a-=144:
a=a-144=144-144=0
3.
a+=0:
a=a+0=0+0=0
⑽ 求一个A*算法的C语言或C++代码，小弟不胜感激，谢谢
1#include <iostream>
2#include <queue>
3usingnamespace std;
4
5struct knight{
6int x,y,step;
7int g,h,f;
8booloperator< (const knight & k) const{ //重载比较运算符
9return f > k.f;
10 }
11}k;
12bool visited[8][8]; //已访问标记(关闭列表)
13int x1,y1,x2,y2,ans; //起点(x1,y1),终点(x2,y2),最少移动次数ans
14int dirs[8][2]={{-2,-1},{-2,1},{2,-1},{2,1},{-1,-2},{-1,2},{1,-2},{1,2}};//8个移动方向
15priority_queue<knight> que; //最小优先级队列(开启列表)
16
17boolin(const knight & a){ //判断knight是否在棋盘内
18if(a.x<0|| a.y<0|| a.x>=8|| a.y>=8)
19returnfalse;
20returntrue;
21}
22int Heuristic(const knight &a){ //manhattan估价函数
23return (abs(a.x-x2)+abs(a.y-y2))*10;
24}
25void Astar(){ //A*算法
26 knight t,s;
27while(!que.empty()){
28 t=que.top(),que.pop(),visited[t.x][t.y]=true;
29if(t.x==x2 && t.y==y2){
30 ans=t.step;
31break;
32 }
33for(int i=0;i<8;i++){
34 s.x=t.x+dirs[i][0],s.y=t.y+dirs[i][1];
35if(in(s) &&!visited[s.x][s.y]){
36 s.g = t.g +23; //23表示根号5乘以10再取其ceil
37 s.h = Heuristic(s);
38 s.f = s.g + s.h;
39 s.step = t.step +1;
40 que.push(s);
41 }
42 }
43 }
44}
45int main(){
46char line[5];
47while(gets(line)){
48 x1=line[0]-'a',y1=line[1]-'1',x2=line[3]-'a',y2=line[4]-'1';
49 memset(visited,false,sizeof(visited));
50 k.x=x1,k.y=y1,k.g=k.step=0,k.h=Heuristic(k),k.f=k.g+k.h;
51while(!que.empty()) que.pop();
52 que.push(k);
53 Astar();
54 printf("To get from %c%c to %c%c takes %d knight moves.\n",line[0],line[1],line[3],line[4],ans);
55 }
56return0;
57}
58