7-1 The Closest Fibonacci Number

最新推荐文章于 2025-12-09 11:28:31 发布

原创最新推荐文章于 2025-12-09 11:28:31 发布 · 215 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#c++ #开发语言

pat 专栏收录该内容

33 篇文章

订阅专栏

该程序实现了一个算法，用于找到给定整数N最接近的斐波那契数。输入是一个不超过10^8的正整数N，输出是最小绝对差的斐波那契数。示例中，当N为305时，最接近的斐波那契数有两个：233和377，两者与305的差值都是72，但程序会输出较小的那个，即233。

The Fibonacci sequence Fn is defined by Fn+2=Fn+1+Fn for n≥0, with F0=0 and F1=1. The closest Fibonacci number is defined as the Fibonacci number with the smallest absolute difference with the given integer N.

Your job is to find the closest Fibonacci number for any given N.

Input Specification:

Each input file contains one test case, which gives a positive integer N (≤108).

Output Specification:

For each case, print the closest Fibonacci number. If the solution is not unique, output the smallest one.

Sample Input:

Sample Output:

Hint:

Since part of the sequence is { 0, 1, 1, 2, 3, 5, 8, 12, 21, 34, 55, 89, 144, 233, 377, 610, ... }, there are two solutions: 233 and 377, both have the smallest distance 72 to 305. The smaller one must be printed out.

#include<iostream>
using namespace std;
const int N = 1000010;
int f[N];
int fib(int n){
    f[0] = 0,f[1] = 1;
    int i = 1;
    for(i = 2;;i++){
        f[i] = f[i-1]+f[i-2];
        if(f[i]>=n)break;
    }
    return i;
}
int main(){
    int n;
    scanf("%d",&n);
    int k = fib(n);
    int n1 = 0x3f3f3f3f,n2 = 0x3f3f3f3f;
    if(k>0) n1 = abs(f[k-1]-n);
    n2 = abs(f[k]-n);
    if(n1<=n2) printf("%d",f[k-1]);
    else printf("%d",f[k]);
}