topic
[Fibonacci Number] The Fibonacci number series is defined as this:
f [0] = 0
f [1] = 1
for EACH I ≥ 2: f [i] = f [i-1] + f [i-2]
. We call the Fibonacci number. Give you an N, you want to change it into a Fibonacci number, you can change the current number X into X-1 or X+1 at each step
, now give you a number N and find the minimum number of steps needed to become a Fibonacci number.
Enter a description:
The input is a positive integer N(1 ≤ N ≤ 1,000,000)
Output description:
Output a minimum number of steps into Fibonacci number"
Example 1:
Enter:
15
output:
2
Answer
#define _CRT_SECURE_NO_WARNINGS 1
#include<stdio.h>
#include<math.h>
int main()
{
int n = 0;
scanf("%d", &n);
int f1 = 0;
int f2 = 1;
int f3 = f2 + f1;
while (1)
{
if (n == f2)
{
printf("0\n"); //因为输入的n大于等于1，所以n也可能一开始等于f2，n=f2就只需要0步变换
}
else if (n < f2)
{
if (abs(n - f2) < abs(n - f1)) //abs为求绝对值函数
{
printf("%d\n", abs(n - f2));
}
else
{
printf("%d\n", abs(n - f1));
}
}
f1 = f2; //n大于f2，循环会一直执行下去，直到n处于f1跟f2之间，随后做差求出最小的步数
f2 = f3;
f3 = f1 + f2;
}
return 0;
}
first contact ideas
#include<stdio.h>
int main()
{
int n=0;
int a=1;
int i=0;
int num=1;
int arr[10000]={
0};
int j=0;
int y=0;
scanf("%d",&n);
for(i=0;num<=1000000;i=y) //Fibonacci数列一个一个找出来放入数组
{
y=num;
num+=i;
arr[j]=num;
j++;
}
int c=0;
for(c=0;c<j-1;c++) //遍历数组对比
if((arr[c]<=n)&&(arr[c+1]>=n))
{
printf("%d\n",((n-arr[c])>(arr[c+1]-n))?(arr[c+1]-n):(n-arr[c]));
}
return 0;
}
First find out the Fibonacci sequence one by one, and then traverse the arrays and compare them one by one, which is indeed blind and troublesome.