再来看看这一个测试题

编程输出斐波那契数列的前n项。隻波那契数列是前两项为1，从第3项开始，每项都是前两项之和。输入正整数n (n<=30)，输出斐波那契数列的前n项。项之间用一个空格分隔，最后一项之后也有一个空格，
输入
10
输出

#include <stdio.h>
 
int main()
{
    int i, n, t1 = 1, t2 = 1, nextTerm;
 
    printf("输出几项: ");
    scanf("%d", &n);
 
    printf("斐波那契数列: ");
 
    for (i = 1; i <= n; ++i)
    {
        printf("%d ", t1);
        nextTerm = t1 + t2;
        t1 = t2;
        t2 = nextTerm;
    }
    return 0;
}

#include <stdio.h>
// 斐波那契的递归函数
int fbi(int i)
{
if( i < 2 )
return i == 0 ? 0 : 1;
return fbi(i - 1) + fbi(i - 2);
}

int main()
{
int n;
scanf("%d",&n);
int i;
for(i = 0;i <= n;i++)
{
printf("%d ", fbi(i));
}
printf("\n");
return 0;
}