不可以，因为函数返回值具有唯一性，一元二次方程的根不能设置成以函数返回值的方式返回给主调函数，，
`#include <stdio.h>
long long fac(int n);
int main() {
int i,n;
printf("Enter n: ");
scanf("%d", &n);
for(i=1; i<=n; ++i)
printf("%d! = %lld\n", i, fac(i));
return 0;
}
long long fac(int n) {

static long long p = 1;
p = p*n;
return p;

}`

`#include <stdio.h>
long long fac(int n);
int main() {
int i,n;
printf("Enter n: ");
scanf("%d", &n);
for(i=1; i<=n; ++i)
printf("%d! = %lld\n", i, fac(i));
return 0;
}
long long fac(int n)
{
static long long p = 1;
printf("p=%11d\n",p);
p = p*n;

return p;

}`

#include<stdio.h> int func(int, int); int main() { int k=4,m=1,p1,p2; p1 = func(k,m) ; p2 = func(k,m) ; printf("%d,%d\n",p1,p2) ; return 0; } int func(int a,int b) { static int m=0,i=2; i += m+1; m = i+a+b; return (m); }

`#include <stdio.h>

define N 1000

int fun(int n,int m,int bb[N]) {
int i,j,k=0,flag;
for(j=n;j<=m;j++) {
flag=1;
for(i=2;i<j;i++)
if(j%i0)
{
flag=0;
break;
}
if(ji)
bb[k++]=j;
}
return k;
}
int main(){
int n=0,m=0,i,k,bb[N];
scanf("%d",&n);
scanf("%d",&m);
for(i=0;i<m-n;i++)
bb[i]=0;
k=fun(n,m,bb);
for(i=0;i<k;i++)
printf("%4d",bb[i]);

return 0;
}`

#include <stdio.h> long long fun(int n); int main() { int n; long long f; while(scanf("%d", &n) != EOF) { f = fun(n); printf("n = %d, f = %lld\n", n, f); } return 0; } long long fun(int n) { long long y; if(n==0) return 0; else y=2*fun(n-1)+1; return y; }

`#include <stdio.h>
void draw(int n, char symbol);

include <stdio.h>

int main()
{
int n, symbol;
while(scanf("%d %c", &n, &symbol) != EOF)
{
draw(n, symbol);
printf("\n");
}
return 0;
}
void draw(int n,char symbol)
{
int i,j;
for(i=0;i<n;i++)
{
for(j=0;j<n-i-1;j++)
printf(" ");
for(j=0;j<=2*i;j++)
printf("%c",symbol);
printf("\n");
}
}`