任務一

每個函式只有一個返回值，一元二次方程的根有兩個，故不能設計成返回值。

任務二

#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>
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);
}

任務三

#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%i==0){
flag=0;
break;
}
if(flag==1)
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;
}

任務四

#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 s;
    if(n==0)
      s=0;
    else
      s=2*fun(n-1)+1;
      return s;
}

任務五

#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 line,i;
    for(line=1;line<=n;line++){
    for(i=1;i<=n-line;i++)
    printf(" ") ;
    for(i=1;i<=2*line-1;i++)
    printf("%c",symbol);
    printf("\n");
    }
}