Question

用C语言编写21个斐波那契书的和的程序

Answer 1

#include<stdio.h>
void main()
{ int i;
long f1=1,f2=1;
printf("前21组菲薄纳妾数列为\n");
for(i=0;i<21;i++)/*共输21组数*/
{printf("%12ld%12ld",f1,f2);
f1=f1+f2;
f2=f2+f1;
}
}

Answer 2

#include<stdio.h>
#define N 21

main(){
int i,a[N],s=0;
char c;
for(i=0;i<N;i++){
if(i<2) a[i]=1;
else
a[i]=a[i-1]+a[i-2];
}
for(i=0;i<N;i++){
printf("%d,",a[i]);s+=a[i];}
printf("是否求和？(Y/N)\n");
c=getch();
if(c=='y'||c=='Y')
printf("\n\ts=%d",s);
getch();
}

Answer 3

主函数
main()
{int a[21],i;
a[2]=1;
a[1]=1;
for (i=2;i<21;i++)
a[i]=a[i-1]+a[i-2];
for (i=1;i<21;i++)
printf ("%d ",a[i]);
}