斐波那契数列第一种#include<stdio.h>#define n 10 //一共n/2行void main(){int f,g,i;f=1;g=1;for(i=1;i<=n;i++) { printf("%10d%10d",f,g); if(i%2==0) printf("\n");//控制每行输出4项 f=f+g;g=g+f; }}第二种#include<stdio.h>void main(){int i,a[10]={1,1};for(i=2;i<10;i++)a[i]=a[i-1]+a[i-2];for(i=0;i<10;i++){if(i%5==0) printf("\n");printf("%5d",a[i]);}}