PAT_甲级_2020年冬季考试 7-1 The Closest Fibonacci Number

7-1 The Closest Fibonacci Number (20 分)

The Fibonacci sequence Fn is defined by $$F_{n+2} =F_{n+1} +F​_n \;\;for \;\;n≥0, with \;\;F_​0 =0 \;\;and\;\; F_​1 =1.$$ 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 (≤10^8).

Output Specification:

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

Sample Input:

305

Sample Output:

233

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.

题目要求

题目大意

给定一个N，要求输出斐波拉契数列中最接近N的数字。

算法思路

需要不断生成斐波拉契数列的每一项，遇到第一次大于等于N的数字就停止，然后记录当前位置为k，判断k-1和k位置的那个数字和N更接近就输出哪个。生成斐波那契数列的方法就是采用递归加上记忆化搜索(使用dp数组作为一个缓存，遇到算过的值就直接返回，否则就递归计算，然后保存)

注意点

• 1、直接暴力递归会导致测试点5运行超时

提交结果

AC代码

#include<cstdio>
#include<algorithm>
#include<cstring>

using namespace std;

int dp[1000000];

int F(int n) {
    if (n == 0 || n == 1) {
        dp[n] = n;
        return n;
    }
    if(dp[n]!=-1){
        return dp[n];
    }else{
        dp[n] = F(n - 1) + F(n - 2);
    }
    return dp[n];
}

int main() {
    int N;
    scanf("%d", &N);
    memset(dp, -1, sizeof(dp));
    dp[0] = dp[1] = 1;
    int k;
    for (int i = 0;; ++i) {
        if (dp[i] == -1) {
            dp[i] = F(i);
        }
        if (dp[i] >= N) {
            k = i;
            break;
        }
    }
    int a = abs(dp[k] - N);
    int b = abs(dp[k - 1] - N);
    if (a < b) {
        printf("%d", dp[k]);
    } else {
        printf("%d", dp[k - 1]);
    }
    return 0;
}