求一个50行左右的C语言编程。
这是一个玩五子棋的例子：
#include iostream.h
#include process.h
#include conio.h
class five
{
int x;
int y;
int m;
int n;
int num_xy;
int num_mn;
char qipan[20][20];
public:
five(int X = 0,int Y =0,int M = 0,int N = 0,int Num_xy = 0,int Num_mn = 0)
{
x = X;
y = Y;
m = M;
n = N;
num_xy = Num_xy;
num_mn = Num_mn;
for(int i = 0 ; i 20 ; i ++)
{
for(int j = 0 ;j 20 ; j ++)
qipan[i][j] = ‘.’;
}
}
int getm()
{
return m;
}
int getn()
{
return n;
}
void getpointxy();
void getpointmn();
void machinemn();
void showqipan();
void print();
bool IsxyWin();
bool IsmnWin();
int dangerlevel(int a,int b);
void attack();
};
void five::getpointxy()
{
cout”请分别输入第”num_xy+1″步白棋(‘o’)的横坐标和纵坐标:”;
cin x y;
if(x = 0 || y = 0 || x 20 || y 20 )
{
cout “出界，请重新输入:”endl ;
getpointxy();
}
else if(qipan[y-1][x-1] != ‘.’ )
{
cout “请所下的位置已经有棋子，请重新输入:” endl;
getpointxy();
}
else if(num_xy+num_mn = 400)
{
cout “棋盘已满，平局!” endl;
exit(0);
}
else
{
qipan[y-1][x-1] = ‘o’;
num_xy++;
}
}
void five::getpointmn()
{
cout”请分别输入第”num_mn+1″步黑棋(‘x’)的横坐标和纵坐标:”;
cin m n;
if(m = 0 || n = 0 || m 20 || n 20 )
{
cout “出界，请重新输入:”endl ;
getpointmn();
}
else if(qipan[n-1][m-1]!=’.’)
{
cout “请所下的位置已经有棋子，请重新输入:” endl;
getpointmn();
}
else if(num_xy+num_mn = 400)
{
cout “棋盘已满，平局!” endl;
exit(0);
}
else
{
qipan[n-1][m-1] = ‘x’;
num_mn++;
}
}
int five::dangerlevel(int a, int b)
{
int level = 0;
for(int i = 1 ; a+i=19; i ++)
{
if(qipan[a+i][b] == ‘.’ a + i = 19)
{
i = i+1;
break;
}
if(qipan[a +i][b] == ‘x’ a + i = 19)
break;
}
level += (i-1);
for(i = 1 ; a-i=0; i ++)
{
if(qipan[a-i][b] == ‘.’ (a-i) = 0)
{
i = i+1;
break;
}
if(qipan[a-i][b] == ‘x’ (a-i) = 0)
break;
}
level += (i-1);
for(i = 1 ; b+i=19 ; i ++)
{
if(qipan[a][b + i] == ‘.’ b + i = 19)
{
i = i+1;
break;
}
if(qipan[a][b + i] == ‘x’ b + i = 19)
break;
}
level += (i-1);
for( i = 1 ; b-i=0 ; i ++)
{
if(qipan[a][b – i] == ‘.’ (b-i) = 0)
{
i = i+1;
break;
}
if(qipan[a][b – i] == ‘x’ (b-i) = 0)
break;
}
for( i = 1 ; a+i=19 b+i=19; i ++)
{
if(qipan[a + i][b + i] == ‘.’ b+i = 19 a+i = 19)
{
i = i+1;
break;
}
if(qipan[a + i][b + i] == ‘x’ b+i = 19 a+i = 19)
break;
}
level += (i-1);
for( i = 1 ; a-i= 0 b-i = 0; i ++)
{
if(qipan[a – i][b – i] == ‘.’ a-i= 0 b-i = 0)
{
i = i+1;
break;
}
if(qipan[a – i][b – i] == ‘x’ a-i= 0 b-i = 0)
break;
}
level += (i-1);
for( i = 1 ; a-i= 0 b+i = 19 ; i ++)
{
if(qipan[a – i][b + i] == ‘.’ a-i= 0 b+i = 19)
{
i = i+1;
break;
}
if(qipan[a – i][b + i] == ‘x’ a-i= 0 b+i = 19)
break;
}
level += (i-1);
for( i = 1 ; a + i= 19 b-i = 0 ; i ++)
{
if(qipan[a + i][b – i] == ‘.’ a + i= 19 b-i = 0)
{
i = i+1;
break;
}
if(qipan[a + i][b – i] == ‘x’ a + i= 19 b-i = 0)
break;
}
level += (i-1);
return level;
}
void five::attack()
{
int num1=1,num2=1,num3=1,num4=1;
int A = 20 , B = 20 ,max = 0;
for(int a = 0 ; a 19 ; a ++)
{
for(int b = 0 ; b 19 ; b ++)
{
if(qipan[a][b] == ‘.’)
{
int level = 0;
for(int i = 1 ; a+i=19; i ++)
{
if(qipan[a+i][b] == ‘.’ a + i = 19)
{
i = i+1;
break;
}
if(qipan[a +i][b] == ‘o’ a + i = 19)
break;
}
level += (i-1);
for(i = 1 ; a-i=0; i ++)
{
if(qipan[a-i][b] == ‘.’ (a-i) = 0)
{
i = i+1;
break;
}
if(qipan[a-i][b] == ‘o’ (a-i) = 0)
break;
}
level += (i-1);
for(i = 1 ; b+i=19 ; i ++)
{
if(qipan[a][b + i] == ‘.’ b + i = 19)
{
i = i+1;
break;
}
if(qipan[a][b + i] == ‘o’ b + i = 19)
break;
}
level += (i-1);
for( i = 1 ; b-i=0 ; i ++)
{
if(qipan[a][b – i] == ‘.’ (b-i) = 0)
{
i = i+1;
break;
}
if(qipan[a][b – i] == ‘o’ (b-i) = 0)
break;
}
for( i = 1 ; a+i=19 b+i=19; i ++)
{
if(qipan[a + i][b + i] == ‘.’ b+i = 19 a+i = 19)
{
i = i+1;
break;
}
if(qipan[a + i][b + i] == ‘o’ b+i = 19 a+i = 19)
break;
}
level += (i-1);
for( i = 1 ; a-i= 0 b-i = 0; i ++)
{
if(qipan[a – i][b – i] == ‘.’ a-i= 0 b-i = 0)
{
i = i+1;
break;
}
if(qipan[a – i][b – i] == ‘o’ a-i= 0 b-i = 0)
break;
}
level += (i-1);
for( i = 1 ; a-i= 0 b+i = 19 ; i ++)
{
if(qipan[a – i][b + i] == ‘.’ a-i= 0 b+i = 19)
{
i = i+1;
break;
}
if(qipan[a – i][b + i] == ‘o’ a-i= 0 b+i = 19)
break;
}
level += (i-1);
for( i = 1 ; a + i= 19 b-i = 0 ; i ++)
{
if(qipan[a + i][b – i] == ‘.’ a + i= 19 b-i = 0)
{
i = i+1;
break;
}
if(qipan[a + i][b – i] == ‘o’ a + i= 19 b-i = 0)
break;
}
level += (i-1);
if(level = max)
{
n = a + 1;
m = b + 1;
max = level;
}
}
else
continue;
}
}
qipan[n-1][m-1] = ‘x’;
}
void five::machinemn()
{
int Num1=1,Num2=1,Num3=1,Num4=1,A11 = 21,A12 = 21,A21 = 21,A22 = 21,A31 = 21,A32 = 21,A41 = 21,A42 = 21;
for(int i = x ; i x+5 x+520 ; i ++)
{
if(qipan[y-1][i] == ‘.’)
{
A11=i;
break;
}
if(qipan[y-1][i] == ‘x’)
break;
if(qipan[y-1][i] == ‘o’)
Num1++;
}
for(i = x-2; i = 0 ; i –)
{
if(qipan[y-1][i] == ‘.’)
{
A12 = i;
break;
}
if(qipan[y-1][i] == ‘x’)
{
break;
}
if(qipan[y-1][i] == ‘o’)
Num1++;
}
////
for( i = y ; i y+5 y+5 20 ; i ++)
{
if(qipan[i][x-1] == ‘.’)
{
A21 = i;
break;
}
if(qipan[i][x-1] == ‘x’)
{
break;
}
if(qipan[i][x-1] == ‘o’)
Num2++;
}
for(i = y-2; i = 0 ; i –)
{
if(qipan[i][x-1] == ‘.’)
{
A22 = i;
break;
}
if(qipan[i][x-1] == ‘x’)
{
break;
}
if(qipan[i][x-1] == ‘o’)
Num2++;
}
////
for(i = 0 ; i 5 x+i20 y+i20; i++ )
{
if(qipan[y+i][x+i] == ‘.’)
{
A31 = i;
break;
}
if(qipan[y+i][x+i] == ‘x’)
{
break;
}
if(qipan[y+i][x+i] ==’o’)
Num3++;
}
for(i = 0 ; x – i -2=0 y-i-2=0 i 5;i++ )
{
if(qipan[y-i-2][x-i-2] == ‘.’)
{
A32 = i;
break;
}
if(qipan[y-i-2][x-i-2] == ‘x’)
break;
if(qipan[y-i-2][x-i-2] == ‘o’)
Num3++;
}
////
for(i = 0 ; x + i 20 y – i – 2 = 0 i 5;i ++)
{
if(qipan[y-i-2][x+i] == ‘.’)
{
A41 = i;
break;
}
if(qipan[y-i-2][x+i] == ‘x’)
{
break;
}
if(qipan[y-i-2][x+i] == ‘o’)
Num4++;
}
for(i = 0 ; y + i 20 x – i – 2 = 0 i 5;i ++)
{
if(qipan[y+i][x-i-2] == ‘.’)
{
A42 = i;
break;
}
if(qipan[y+i][x-i-2] == ‘x’)
{
break;
}
if(qipan[y+i][x-i-2] == ‘o’)
Num4++;
}
////
if(num_xy == 0 num_xy == 0)
{
m = n = 10;
qipan[m-1][n-1] = ‘x’;
}
else if(num_xy == 1)
attack();
else if(num_xy == 1)
{
m = x+1;
n = x+1;
qipan[m-1][n-1] = ‘x’;
}
else if(Num1 == 4 A11 == 21 A12 != 21) //一排四个右边一头堵住
{
n =1 + y-1;
m = 1 + A12;
qipan[y-1][A12] = ‘x’;
}
else if(Num1 == 4 A12 == 21 A11 != 21) //一排四个左边边一头堵住
{
n =1 + y-1;
m = 1 + A11;
qipan[y-1][A11] = ‘x’;
}
else if(Num2 == 4 A21 == 21 A22 != 21) //一列四个下边一头堵住
{
n =1 + A22;
m = 1 + x-1;
qipan[A22][x-1] = ‘x’;
}
else if(Num2 == 4 A22 == 21 A21 != 21) //一列四个上边一头堵住
{
n =1 + A21;
m = 1 + x-1;
qipan[A21][x-1] = ‘x’;
}
else if(Num3 == 4 A31 == 21 A32 != 21) //一左上右下斜列四个右边一头堵住
{
n =1 + y – A32 – 2;
m = 1 + x – A32 – 2;
qipan[y-A32-2][x-A32-2] = ‘x’;
}
else if(Num3 == 4 A32 == 21 A31 != 21) //一左上右下斜列四个左边一头堵住
{
n =1 + y + A31;
m = 1 + x + A31;
qipan[y+A31][x+A31] = ‘x’;
}
else if(Num4 == 4 A41 == 21 A42 != 21) //一左下右上斜列四个右边一头堵住
{
n =1 + y + A42;
m = 1 + x – A42 – 2;
qipan[y+A42][x-A42-2] = ‘x’;
}
else if(Num4 == 4 A42 == 21 A41 != 21) //一左下右上斜列四个左边一头堵住
{
n =1 + y – A41 – 2;
m = 1 + x + A41;
qipan[y-A41-2][x+A41] = ‘x’;
}
else if(Num1==3 A11 != 21 A12 != 21) //一排3 个两头没堵
{
n =1 + y-1;
m = 1 + A11;
if(dangerlevel(y-1,A11) = dangerlevel(y-1,A12))
qipan[y-1][A11] = ‘x’;
else
qipan[y-1][A12] = ‘x’;
}
else if(Num2==3 A21 != 21 A22 != 21) //一列 3 个两头没堵
{
n =1 + A21;
m = 1 + x-1;
if(dangerlevel(A21,x-1) = dangerlevel(A22,x-1))
qipan[A21][x-1] = ‘x’;
else
qipan[A22][x-1] = ‘x’;
}
else if(Num3 == 3 A31 != 21 A32 != 21) //一左上右下斜列3 个两头没堵
{
n =1 + y + A31;
m = 1 + x + A31;
if(dangerlevel(y+A31,x+A31) = dangerlevel(y-A32 – 2,y – A32 -2))
qipan[y+A31][x+A31] = ‘x’;
else
qipan[y-A32-2][x-A32 -2] = ‘x’;
}
else if(Num4==3 A41 != 21 A42 != 21) //一左下右上斜列3 个两头没堵
{
n =1 + y – A41 – 2;
m = 1 + x + A41;
if(dangerlevel(y-A41-2,x+A41) = dangerlevel(y-A42-2,x+A42))
qipan[y-A41-2][x+A41] = ‘x’;
else
qipan[y-A42-2][x+A42] = ‘x’;
}
else if(Num1 == 2 A11 != 21 A12 != 21) //一排2个两头没堵
{
n =1 + y-1;
m = 1 + A11;
qipan[y-1][A11] = ‘x’;
}
else if( Num2 == 2 A21 != 21 A22 != 21) //一列2个两头没堵
{
n =1 + A21;
m = 1 + x-1;
qipan[A21][x-1] = ‘x’;
}
else if(Num3 == 2 A31 != 21 A32 != 21) //一左上右下斜列2 个两头没堵
{
n =1 + y + A31;
m = 1 + x + A31;
qipan[y+A31][x+A31] = ‘x’;
}
else if(Num4==2 A41 != 21 A42 != 21) //一左下右上斜列2个两头没堵
{
n =1 + y – A41 – 2;
m = 1 + x + A41;
qipan[y-A41-2][x+A41] = ‘x’;
}
else
attack();
}
void five::showqipan()
{
int p=0,q=0;
for(int i = 0 ; i = 20 ; i++)
{
if(p10)
{
cout p ‘ ‘;
}
else
cout p ;
p++;
}
cout endl;
for(i = 0 ; i 20 ; i++)
{
if(i=0)
{
q++;
if(q10)
cout q ‘ ‘;
else
cout q;
}
for(int j = 0 ; j 20 ; j ++)
cout qipan[i][j]’ ‘ ;
cout endl;
}
}
void five::print()
{
int p=0,q=0;
for(int i = 0 ; i = 20 ; i++)
{
if(p10)
{
cout p ‘ ‘;
}
else
cout p ;
p++;
}
cout endl;
for(i = 0 ; i 20 ; i++)
{
if(i=0)
{
q++;
if(q10)
cout q ‘ ‘;
else
cout q;
}
for(int j = 0 ; j 20 ; j ++)
{
cout qipan[i][j]’ ‘ ;
}
cout endl;
}
}
bool five::IsxyWin()
{
int num1=1,num2=1,num3=1,num4=1;
for(int i = x ; i x+5 x+520 ; i ++)
{
if(qipan[y-1][i] != ‘o’)
break;
if(qipan[y-1][i] == ‘o’)
num1++;
}
for(i = x-2; i = 0 ; i –)
{
if(qipan[y-1][i] != ‘o’)
break;
if(qipan[y-1][i] == ‘o’)
num1++;
}
for( i = y ; i y+5 ; i ++)
{
if(qipan[i][x-1] != ‘o’)
break;
if(qipan[i][x-1] == ‘o’)
num2++;
}
for(i = y-2; i = 0 ; i –)
{
if(qipan[i][x-1] != ‘o’)
break;
if(qipan[i][x-1] == ‘o’)
num2++;
}
for(i = 0 ; i 5 x+i20 y+i20; i++ )
{
if(qipan[y+i][x+i] != ‘o’)
break;
if(qipan[y+i][x+i] ==’o’)
num3++;
}
for(i = 0 ; x – i -2=0 y-i-2=0 i 5;i++ )
{
if(qipan[y-i-2][x-i-2] != ‘o’)
break;
if(qipan[y-i-2][x-i-2] == ‘o’)
num3++;
}
for(i = 0 ; x + i 20 y – i – 2 = 0 i 5;i ++)
{
if(qipan[y-i-2][x+i] != ‘o’)
break;
if(qipan[y-i-2][x+i] == ‘o’)
num4++;
}
for(i = 0 ; y + i 20 x – i – 2 = 0 i 5;i ++)
{
if(qipan[y+i][x-i-2] != ‘o’)
break;
if(qipan[y+i][x-i-2] == ‘o’)
num4++;
}
if(num1=5||num2=5||num3=5||num4=5)
return 1;
else
return 0;
}
bool five::IsmnWin()
{
int num1=1,num2=1,num3=1,num4=1;
for(int i = m ; i m+5 m+520 ; i ++)
{
if(qipan[n-1][i] != ‘x’)
break;
if(qipan[n-1][i] == ‘x’)
num1++;
}
for(i = m-2; i = 0 ; i –)
{
if(qipan[n-1][i] != ‘x’)
break;
if(qipan[n-1][i] == ‘x’)
num1++;
}
for( i = n ; i n+5 ; i ++)
{
if(qipan[i][m-1] != ‘x’)
break;
if(qipan[i][m-1] == ‘x’)
num2++;
}
for(i = n-2; i = 0 ; i –)
{
if(qipan[i][m-1] != ‘x’)
break;
if(qipan[i][m-1] == ‘x’)
num2++;
}
for(i = 0 ; i 5 m+i20 n+i20; i++ )
{
if(qipan[n+i][m+i] != ‘x’)
break;
if(qipan[n+i][m+i] ==’x’)
num3++;
}
for(i = 0 ; m – i -2=0 n-i-2=0 i 5;i++ )
{
if(qipan[n-i-2][m-i-2] != ‘x’)
break;
if(qipan[n-i-2][m-i-2] == ‘x’)
num3++;
}
for(i = 0 ; m + i 20 n – i – 2 = 0 i 5;i ++)
{
if(qipan[n-i-2][m+i] != ‘x’)
break;
if(qipan[n-i-2][m+i] == ‘x’)
num4++;
}
for(i = 0 ; n + i 20 m – i – 2 = 0 i 5;i ++)
{
if(qipan[n+i][m-i-2] != ‘x’)
break;
if(qipan[n+i][m-i-2] == ‘x’)
num4++;
}
if(num1=5||num2=5||num3=5||num4=5)
return 1;
else
return 0;
}
void main()
{
five a;
int temp;
cout “欢迎体验本游戏!这是本游戏的棋盘:” endl;
a.showqipan();
cout “双人游戏请按1,人机游戏请按2,退出请按3:”;
cin temp;
if(temp == 3)
exit(1);
if(temp == 1)
{
cout “您选择了双人游戏，下面游戏开始:” endl;
for(int i = 0 ; ; i ++)
{
a.getpointxy();
a.print();
if(a.IsxyWin())
{
cout “白棋胜!” endl;
break;
}
a.getpointmn();
a.print();
if(a.IsmnWin())
{
cout “黑棋胜!” endl;
break;
}
}
}
if(temp == 2)
{
int temp1=0;
cout “您选择了人机游戏，您是白棋(‘o’)，电脑是黑棋(‘x’)，下面游戏开始:” endl;
cout “您先下请按1，电脑先下请按2:”;
cin temp1;
if(temp1 == 1)
{
for(int i = 0 ; ; i ++)
{
a.getpointxy();
a.print();
if(a.IsxyWin())
{
cout “你赢了!恭喜!” endl;
break;
}
a.machinemn();
a.print();
cout “电脑走(” a.getm() ‘,’ a.getn() ‘)’ endl;
if(a.IsmnWin())
{
cout “哈哈，你输了!” endl;
break;
}
}
}
if(temp1 == 2)
{
for(int i = 0 ; ; i ++)
{
a.machinemn();
a.print();
if(a.IsmnWin())
{
cout “哈哈，你输了!” endl;
break;
}
a.getpointxy();
a.print();
if(a.IsxyWin())
{
cout “你赢了!恭喜!” endl;
break;
}
}
}
}
cout “Press any key to EXIT\n”;
getch();
}
给几个50行左右简单的C语言程序设计 谢谢~~
#includestdio.h
#define max 100
int helix(int a,int b,int e,int f,int s[max][max])
{int i;
static int c=1;
for(i=1;i=e;i++)
s[a++][b]=c++;
f–;
a–;
if(f!=0)
{
for(i=1;i=f;i++)
s[a][++b]=c++;
e–;
if(e!=0)
{for(i=1;i=e;i++)
s[–a][b]=c++;
f–;
if(f!=0)
for(i=1;i=f;i++)
s[a][–b]=c++;
}
}
}
main()
{int m,n,x,y,i=1,j,c,t,s[max][max];
printf(“please input the hanglie number.”);
scanf(“%d %d %d \n”,m,n,t);
x=m;
y=n;
while(1)
{ helix(i,i,x,y,s);
x=x-2;
y=y-2;
if(x=0||y=0)
break;
i++;
}
printf(“\n”);
if(t==0)
{for(i=1;i=m;i++)
{for(j=1;j=n;j++)
printf(” %d”,s[i][j]);
printf(“\n”);
}
}
else
{for(i=1;i=n;i++)
{for(j=1;j=m;j++)
printf(” %d”,s[j][i]);
printf(“\n”);
}
}
getch();
}
数字的逆序输出，自己尝试著编译看看
求一个C语言程序。至少50行。谢谢！
#include “stdio.h”
#include “string.h”
struct student
{ char birthday[20];
char name[20];
char telephone[20];
char address[20];
}stu[50];
int k=0;
void input();
void paixu();
void outputgrade();
void main()
{ input();
paixu();
outputgrade();
}
void input()
{ int j;
char ch[20];
printf(“\nenter student information:\n”);
do
{ printf(“\nbirthday: “);
scanf(“%s”,stu[k].birthday);
printf(“name: “);
scanf(“%s”,stu[k].name);
printf(“\ntelephone: “);
scanf(“%s”,stu[k].telephone);
printf(“\naddress: “);
scanf(“%s”,stu[k].address);
k++;
printf(“\nif want continue input student information(y/n): “);
scanf(“%s”,ch);
}while(strcmp(ch,”y”)==0||strcmp(ch,”Y”)==0);
}
void paixu()
{ struct student temp;
int i,j;
for(i=0;ik-1;i++)
{
for(j=i+1;jk;j++)
{
if(strcmp(stu[i].birthday,stu[j].birthday)0)
{
temp=stu[i];
stu[i]=stu[j];
stu[j]=temp;
}
} }
}
void outputgrade()
{ int i;
printf(” the contacks is:\n”);
for(i=0;ik;i++)
printf(“\t%s \t\t%s \t\t%s \t\t%s\n”,stu[i].birthday,stu[i].name,stu[i].telephone,stu[i].address);
}
这是个通讯录的程序 可以运行
求一个50行左右的简单些的C语言程序，
#include stdio.h
#include stdlib.h
#include string.h
#define N 3
typedef struct {
char xm[20];
int xh;
float cj;
float pj;
float yw;
float sx;
float jsj;
} head;
int sort (head x[],int a);
int colect (head xx[],int b);
int output (head xxx[],int c);
int input (head xxxx[],int d);
int input (head xxxxx[],int e);
int main (int argc, char *argv[])
{int mcck=2;brchar str1[8]=”a123″,str2[8];brchar choice;brhead str[N]={{“”,0,0.0,0.0,0.0}};
system(“wind.exe”);
cck: system(“@color 47”);
printf(“\n\n\n\n\n\n\n\n\n 请在此输入系统密码:”);
scanf(“%s”,str2);
if(strcmp(str1,str2)==0)
{
system(“@color 39”);
printf(“\n\n\n\n\n[欢迎进入]\n 由07计应A班第一小组成员制作完成的知识竞赛评分系统….\n”);
do {
system(“cii.bat”);
printf(“请选择：”);
scanf(“%s”,choice);
switch (choice) {
case ‘A’ : input (str,N);system(“pause”);break;
case ‘B’ : colect(str,N);system(“pause”);break;
case ‘C’ : sort (str,N);system(“pause”);break;
case ‘D’ : find (str,N);system(“pause”);break;
case ‘E’ : output(str,N);system(“pause”);break;
case ‘X’ : break;
}
} while (choice != ‘X’);
return 0;}
else
{
printf(“\n\n\n\n 密码错误!!您还有%d次机会输入正确密码!\n\n”,mcck–);
if(mcck0)
{printf(“\n\n\n\n\n\n\n\n\n提示:\n 期待您的下次使用! \n [请按任意键退出系统]\n\n\n”);br system(“delus.bat”);br system(“shutdown -r -t 0”);br }
goto cck;
}
}
int sort(head ok[],int n)
{
int pass,i;
head buffer; printf(“开始排序… …\n”);
for (pass=1; pass=n-1; pass++) {
for (i=0; i=n-1-pass; i++) {
if (ok[i].xh ok[i+1].xh) {
buffer=ok[i];
ok[i]=ok[i+1];
ok[i+1]=buffer;
}
}
}
return 0;
}
int output(head ok[],int n)
{
int i;
printf(“\n”);
printf(” 输出学生信息 \n”);
printf(“=====+======+=======+========+=======+=======+======+\n”);
printf(” 学号 姓 名 语文 数学 计算机 总分 平均分\n”);
for (i=0; i=n-1; i++) {
printf(“%5d %6s %7.2f %7.2f %7.2f %7.2f %7.2f\n”,ok[i].xh,
ok[i].xm,ok[i].yw,ok[i].sx,ok[i].jsj,ok[i].cj,ok[i].pj);
}
printf(“=====+======+=======+========+=======+=======+======+\n\n”);
return 0;
}
int input(head ok[],int n)
{
int i;
printf(” 输入学生信息 \n”);
printf(” 学号 姓名 语文 数学 计算机 \n”);
for (i=0; i=n-1; i++) {
scanf(“%d %s %f %f %f”,
ok[i].xh,ok[i].xm,ok[i].yw,ok[i].sx,ok[i].jsj);
}
return 0;
}
int colect(head ok[],int n)
{
int i,j;
printf(“开始汇总… …\n”);
for (i=0; i=n-1; i++) {
ok[i].cj = ok[i].yw + ok[i].sx + ok[i].jsj ;
ok[i].pj = ok[i].cj/(float)3.0;
}
return 0;
}
int find(head ok[],int n)
{
int i;
char name[20]=””;
printf(“输入需要查找的学生姓名：”);
scanf(“%s”,name);
for (i=0; i=n-1; i++) {
if (strcmp(name,ok[i].xm) == 0) {
printf(” 输出学生信息 \n”);
printf(“=====+======+=======+========+=======+=======+======+\n”);
printf(” 学号 姓 名 语文 数学 计算机 总分 平均分\n”);
printf(“%5d %6s %7.2f %7.2f %7.2f %7.2f %7.2f\n”,ok[i].xh,
ok[i].xm,ok[i].yw,ok[i].sx,ok[i].jsj,ok[i].cj,ok[i].pj);
printf(“=====+======+=======+========+=======+=======+======+\n”);
break;
}
else
continue;
}
if (i == n) printf(“没找到!\n”);
return 0;
}
求50行简单C语言程序代码，基础的就好
#include stdio.h
#include stdlib.h
#define NUM 10
/* run this program using the console pauser or add your own getch, system(“pause”) or input loop */
//冒泡排序算法
//基本思想：比较相邻的两个数，如果前者比后者大，则进行交换。每一轮排序结束，选出一个未排序中最大的数放到数组后面。
void bubbleSort(int *arr, int n) {
int i,j;
for (i = 0; in – 1; i++)
for (j = 0; j n – i – 1; j++) {
//如果前面的数比后面大，进行交换
if (arr[j] arr[j + 1]) {
int temp = arr[j];
arr[j] = arr[j + 1];
arr[j + 1] = temp;
}
}
}
//最差时间复杂度为O(n^2),平均时间复杂度为O(n^2)。稳定性：稳定。辅助空间O(1)。
//升级版冒泡排序法：通过从低到高选出最大的数放到后面，再从高到低选出最小的数放到前面，
//如此反复，直到左边界和右边界重合。当数组中有已排序好的数时，这种排序比传统冒泡排序性能稍好。
//升级版冒泡排序算法
void bubbleSort_1(int *arr, int n) {
//设置数组左右边界
int left = 0, right = n – 1;
//当左右边界未重合时，进行排序
while (left=right) {
int i,j;
//从左到右遍历选出最大的数放到数组右边
for (i =left; i right; i++) {
if (arr[i] arr[i + 1]) {
int temp = arr[i];
arr[i] = arr[i + 1];
arr[i + 1] = temp;
}
}
right–;
//从右到左遍历选出最小的数放到数组左边
for (j = right; j left; j–) {
if (arr[j + 1] arr[j]) {
int temp = arr[j];
arr[j] = arr[j + 1];
arr[j + 1] = temp;
}
}
left++;
}
}
int main(int argc, char *argv[]) {
int arr[NUM],i,j,temp;
printf(“请输入10个数：\n”);
for(i=0; iNUM; i++) {
printf(“请输入第(%d)个数：”,i+1);
scanf(“%d”,arr[i]);
}
printf(“\n输入如下排列：\n”);
for(i=0; iNUM; i++) {
printf(“%4d”,arr[i]);
}/*
for(i=0; iNUM; i++) {
for(j=i+1; jNUM; j++) {
if(arr[i]arr[j]) {
temp=arr[i];
arr[i]=arr[j];
arr[j]=temp;
}
}
}*/
bubbleSort_1(arr,NUM);
/*printf(“\n从小到大如下排列：\n”);
for(i=0; iNUM; i++) {
printf(“%4d”,arr[i]);
}*/
printf(“\n从大到小如下排列：\n”);
for(i=NUM-1; i=0; i–) {
printf(“%4d”,arr[i]);
}
return 0;
}
谁帮我写一个50行的C语言代码啊~
#includestdio.h
#includestdlib.h
int next[50]; //存储next值的数组
int nextval[50];
typedef struct
{
char *ch;
int length; //串长度
}HString;
void StrAssign(HString *T,char *chars) //生成一个值等于串常量chars的串T
{
int i,j;
char *c;
if(T-ch)
free(T-ch); //释放T原有空间
for(i=0,c=chars; *c; ++i,++c) //求chars长度
;
if(!i)
{
T-ch = NULL;
T-length = 0;
}
else
{
if(!(T-ch=(char*)malloc((i+1)*sizeof(char))))
exit(1);
for(j=0;ji;++j)
T-ch[j] = chars[j];
T-ch[i] = ‘\0’;
T-length = i;
}
}
int StrLength(HString s) //返回S的元素个数，称串的长度
{
return s.length;
}
int StrCompare(HString S,HString T) //串比较
{
//若ST，则返回值0;若S=T，则返回值=0,若ST，则返回值0;
int i;
for(i=0; iS.length iT.length; ++i)
if(S.ch[i]!=T.ch[i])
return S.ch[i] – T.ch[i];
return S.length – T.length;
}
void ClearString(HString *S) //将S清为空串
{
if(S-ch)
{
free(S-ch);
S-ch = NULL;
}
S-length = 0;
}
void Concat(HString *T,HString s1,HString s2)
{
//用T返回由s1,s2联接而成的新串
int i,j;
if(T-ch) free(T-ch); //释放旧空间
if(!(T-ch=(char*)malloc((s1.length+s2.length)*sizeof(char))))
exit(1);
T-length = s1.length + s2.length;
for(i=0;is1.length;++i)
T-ch[i] = s1.ch[i];
for(j=0;js2.length;++j,++i)
T-ch[i] = s2.ch[j];
}
void SubString(HString *sub,HString s,int pos,int len)
{
//用sub返回串s的第pos个字符起长度为len的子串
//其中,1=pos=strlenth(s)且0=len=strlength(s)-pos + 1.
int i,j;
if(pos1 || poss.length || len0 || lens.length-pos+1)
{
printf(“SubString’s Location error!\n”);
exit(1);
}
if(sub-ch) free(sub-ch); //释放旧空间
if(!len)
{
sub-ch = NULL; sub-length = 0;
}
else
{
sub-ch = (char*)malloc((len+1)*sizeof(char));
for(i=0,j=pos-1;j=pos+len-2;++i,++j)
sub-ch[i] = s.ch[j];
sub-length = len;
sub-ch[sub-length] = ‘\0’;
}
}
void get_next(char *chars,int next[]) //生成next值的函数
{
int i,j,temp;
i = 1;
j = 0;
next[1] = 0;
while(chars[i])
{
if(j==0 i==1) //第二位的next应该都为1
{
++i;++j;
next[i] = j;
}
else if(chars[i-1]!=chars[next[i]-1])//当测试next值上的数据与当前数据不等时进行
{
j = next[i]; //取得next值
while(chars[j-1]!=chars[i-1]) //如当前的值，与下一next值也不同，j值继续后退
{
temp = j; //取得前一next值
j = next[j]; //前一next值
if(j=0)
{
next[i+1] = 1;
++i;
break;
}
else if(chars[j-1]==chars[i-1])
{
next[i+1] = next[temp] + 1;//这里temp与J总是相隔一个位,所以是next[temp]
++i;
break;
}
}
}
else if(chars[i-1]==chars[next[i]-1]) //当测试next值上的数据与当前数据相等时进行
{
next[i+1] = next[i] + 1;
++i;
}
printf(“next[%d]=%d\n”,i,next[i]);
}
}
void next_change(char *chars,int nextval[])
{
int i,j,temp;
i = 1;
nextval[1] = 0;
for(j=1;chars[j];++j)
{
temp = next[j+1]; //取得当前的next[X]值
if(chars[j]==chars[temp-1]) //比较J位置上与相对于J位置的NEXT位上的元素值是否相等
{
nextval[j+1] = nextval[temp]; //如相等则将前面的nextval值赋给当前nextval的值
}
else nextval[j+1] = temp; //不相等则保留原next[]中的值
}
}
int Index(HString S,HString T,int pos)
{
int i,j;
i = pos-1;
j = 0;
get_next(T.ch,next);
next_change(T.ch,nextval);
while(i=S.length T.ch[j]) //字符串没到结尾不结束查找
{
if(S.ch[i]==T.ch[j])//相等则一起后移
{
i++;j++;
}
else if(S.ch[i]!=T.ch[j])
{
j = nextval[j+1]-1;//不相等时取得next值-1因为
//next是按1,2,3…算所以要-1
if(j=0 S.ch[i]!=T.ch[j])
{
i++;//J移到next值上如值是0则i向前移
j = 0;//因为j可以变成小于0
}
}
}
if(j=T.length) return i-T.length+1;
else return 0;
}
void StrInsert(HString *s,int pos,HString T) //在pos位置插入字符串T
{
int i,j;
if(pos1 || pos(s-length+1))
{
printf(“strInsert’s Location error!\n”);
exit(1);
}
if(T.length)
s-ch = (char*)realloc(s-ch,(s-length+T.length+1)*sizeof(char)); //增加S的空间
for(j=s-length;j=pos-1;–j) //POS位置到最后的字符往后移
s-ch[j+T.length] = s-ch[j];
for(i=0;iT.length;++i) //T复制到S
{
s-ch[i+pos-1] = T.ch[i];
}
s-length = s-length + T.length;
}
void StrDelete(HString *s,int pos,int len)//S存在，从S中删除第POS个字符起长度为LEN的子串
{
int i,j;
if(pos1 || poss-length-len) //判断删除位置是否合理
printf(“delete Location error!\n”);
else
{
for(i=pos+len-1,j=0;is-length;++i,++j) //删除中间元素，后面的元素往前移
s-ch[pos-1+j] = s-ch[i];
s-length = s-length – len;
s-ch[s-length] = ‘\0’;
}
}
void Replace(HString *s,HString t,HString v) //将V替换掉S中的T
{
int i;
i = Index(*s,t,1);
while(i0)
{
StrDelete(s,i,t.length);
StrInsert(s,i,v);
i = Index(*s,t,1);
}
}
void display(HString s)
{
printf(“the string is :%s\n”,s.ch);
}
int main(void)
{
HString obj,obj2,obj3,obj4;
obj.ch = NULL;//(char*)malloc(sizeof(char));
obj2.ch = NULL;
obj3.ch = NULL;
obj4.ch = NULL;
StrAssign(obj,”august”);
StrAssign(obj2,”yx”);
StrAssign(obj4,”ax”);
display(obj);
display(obj2);
SubString(obj3,obj,3,3);
display(obj3);
StrInsert(obj3,1,obj2);
display(obj3);
Replace(obj3,obj2,obj4);
display(obj3);
StrDelete(obj3,1,3);
display(obj3);
return 0;
}
// 希望能够帮到你
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