以下是引用rjsp在2016-3-21 11:22:19的发言：

换个公式，s(n) = s(n-2) + s(n-1) + 1; s(-1)=0, s(0)=0
#include <stdio.h>

// s(n) = s(n-2) + s(n-1) + 1; s(-1)=0, s(0)=0
void fibonacci_avg( unsigned n )
{
    // 斐波那契数列前n项之和
    unsigned a[6]={0}, b[6]={0};
    for( unsigned m=0; m!=n; ++m )
    {
        unsigned carry = 1;
        for( size_t i=0; i!=6; ++i )
        {
            carry += a + b;
            a = b;
            b = carry%10000000;
            carry /= 10000000;
        }
    }

    // 平均值
    unsigned carry = 0;
    for( size_t i=0; i!=6; ++i )
    {
        carry = carry*10000000 + b[5-i];
        a = carry/n;
        carry %= n;
    }

    // 十进制输出
    printf( "%u%07u%07u%07u%07u%06u%g\n", a[0],a[1],a[2],a[3],a[4],a[5]/10,a[5]%10+carry*1.0/n );

    return;
}

int main( void )
{
    fibonacci_avg( 200 );
    return 0;
}输出