void cnt(char *tt, int pp[])
{

}

main()
{
char tt[1000];
int pp[26], k, n;
clrscr() ;
printf("\nplease enter a char string:") ; scanf("%s",t
t);
cnt(tt,pp);
for(k = 0; k<26; k++) printf ("%d ",pp[k]);
printf("\n");
readwritedat();
}

void readwritedat()
{
char tt[1000];
int pp[26], k, n, i;
FILE *rf, *wf;

rf = fopen("in.dat","r");
wf = fopen("out.dat","w");
for(i = 0; i<10;i++)
{
fscanf(rf,"%s",tt);
cnt(tt,pp);
for(k = 0;k<26; k++) fprintf(wf,"%d",pp[k]);
fprintf(wf,"\n");
}
fclose(rf);
fclose(wf);
}

/* 注：题中必须先对数组pp赋初值。pp[*tt-'a']++的作用是将对应的元素增加1，即
当*tt为'a'时应对pp[0]加1，而pp[*tt-'a']即相当于pp[0]，依此类推。*/
void cnt(char *tt, int pp[])
{
int i;
for(i=0;i<26;i++)  /*数组初始化*/
pp[i]=0;
while(*tt)
{if(*tt>='a'&&*tt<='z')
pp[*tt-'a']++;
tt++;
}
}

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