请用C语言编写一个磁盘读取调度的程序（disk read scheduling）

#include <stdio.h>

#include <string.h>

int main()

{

char s[4]="abc",sc='d'//比较的字符串和字符

FILE *fp

char filename[100],c

int i=0,n1=0,n2=0

long fpos,len

printf("input filename:\n")

gets(filename)//输入文件名称

if((fp=fopen(filename,"r"))==NULL)//打开文件

{

printf("open %s error!\n",filename)

return 1

}

len=strlen(s)

c=fgetc(fp)

while(!feof(fp))

{

if(c==s[0])//如果第一个字符相等,比较剩下的字符串

{

fpos=ftell(fp)//记住当前文件指针位置

for(i=1i<leni++)

{

if(fgetc(fp)!=s[i])//如果不匹配,跳出循环

{

fseek(fp,fpos,0)//重新设置指针位置

break

}

}

if(i==len)//如果匹配成功,累加数目

n1++

}

if(c==sc)//与字符sc匹配,累加数目

n2++

c=fgetc(fp)

}

printf("\n与字符串%s匹配的有%d个\n",s,n1)//输出匹配个数

printf("与字符%c匹配的有%d个\n",sc,n2)

getchar()

return 0

}

前两天刚写完,还没优化,已运行通过了.

晕，一维的好麻烦，这个也是碰巧前两天刚写好的，你看着自己修改下

#include <stdio.h>

typedef struct

{

int line

int row

int num

}Node

int main()

{

/*

int a[9][9]={

{4,0,3,6,0,0,0,0,0},

{0,0,0,0,0,1,0,2,4},

{0,1,0,0,4,0,5,0,0},

{0,0,0,9,0,4,0,6,0},

{3,0,2,0,0,0,4,0,9},

{0,7,4,1,0,3,0,0,0},

{0,0,1,0,9,0,0,4,0},

{2,4,0,3,0,0,0,0,0},

{0,0,0,4,0,8,2,0,7}}

*/

int a[9][9]={

{0,0,0,8,0,0,0,6,0},

{8,7,0,0,0,0,0,0,0},

{2,9,0,0,4,1,0,0,5},

{0,0,5,7,0,0,0,0,9},

{0,2,0,0,0,0,0,1,0},

{9,0,0,0,0,4,3,0,0},

{7,0,0,6,1,0,0,9,8},

{0,0,0,0,0,0,0,5,2},

{0,6,0,0,0,9,0,0,0}}

/*

int a[9][9]={

{0,2,0,0,6,0,0,0,0},

{0,9,0,4,0,5,1,3,0},

{0,0,8,7,0,0,0,0,5},

{6,0,0,3,0,0,4,0,0},

{0,0,0,9,0,6,0,0,0},

{0,0,7,0,0,1,0,0,3},

{4,0,0,0,0,7,3,0,0},

{0,8,5,2,0,4,0,7,0},

{0,0,0,0,9,0,0,1,0}}

*/

/*

int a[9][9]={

{0,0,3,0,2,0,0,0,6},

{0,0,2,0,9,0,0,0,4},

{7,0,0,8,0,0,2,0,3},

{0,8,0,0,7,0,5,0,0},

{0,7,0,1,0,6,0,3,0},

{0,0,0,2,0,0,0,9,0},

{4,0,6,0,0,8,0,0,5},

{6,0,0,0,4,0,3,0,0},

{9,0,0,0,1,0,7,0,0}}

*/

int i,j,n,en,flag,y,k=0,x,qu,p,q

Node b[70]

for(i=0i<9i++)

{

for(j=0j<9j++)

{

if(!a[i][j])

{

b[k].line=i

b[k].row=j

b[k].num=0

k+=1

}

}

}

en=k

/*从b[0]开始试，若b[k].num>9,则k-1,否则k+1*/

for(k=0k<en)

{

++b[k].num

i=b[k].line

j=b[k].row

a[i][j]=b[k].num

n=0

while(n<9&&b[k].num<=9)

{

if(n==i)

{

for(y=0y<9y++)

{

if(y==j)

continue

if(a[n][y]==a[i][j])

flag=1

}

}

else if(n==j)

{

for(y=0y<9y++)

{

if(y==i)

continue

if(a[y][n]==a[i][j])

flag=1

}

}

/*判断同一块中有没有相同值*/

qu=3*(i/3)+j/3

switch(qu)

{

case 0:x=0

y=0

break

case 1:x=0

y=3

break

case 2:x=0

y=6

break

case 3:x=3

y=0

break

case 4:x=3

y=3

break

case 5:x=3

y=6

break

case 6:x=6

y=0

break

case 7:x=6

y=3

break

default :x=6

y=6

break

}

p=x

q=y

for(x<p+3x++)

{

for(y<q+3y++)

{

if(x==i&&y==j)

continue

if(a[x][y]==a[i][j])

{

flag=1

break

}

}

if(flag==1)

break

}

if(flag==1)

{

a[i][j]=++b[k].num

flag=0

n=0

continue

}

n++

}

if(b[k].num>9)

{

a[i][j]=b[k].num=0

k--

if(k<0)

{

printf("error!\r\n")

return -1

}

}

else

k++

}

for(i=0i<9i++)

{

for(j=0j<9j++)

{

printf("%d",a[i][j])

}

printf("\r\n")

}

return 1

}

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