（PAT乙级刷题）最近的斐波那契数

题目：

题解：

#include<iostream>
#include<cmath>
#include <climits>
using namespace std;
int main()
{
    int fib[50] = {0};    //记录10的8次方之内的斐波那契数
    fib[0] = 0, fib[1] = 1;
    int len = 0,i;    //记录斐波那契数的个数
    for (i = 2; fib[i-1] < 1E8; i++)
    {
        fib[i] = fib[i - 1] + fib[i - 2];
    }
    len = i;
    int n;
    cin>>n;
    int min=INT_MAX;
    //找最小值
    for (int i = 0; i < len; i++)
    {
        if(abs(n-fib[i])<min)
        {
            min=abs(n-fib[i]);
        }
    }
    //寻找答案
    for (int i = 0; i < len; i++)
    {
        if(abs(n-fib[i])==min)
        {
            cout<<fib[i];
            return 0;
        }
    }
    return 0;
}