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最新推荐文章于 2024-07-06 00:20:02 发布

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本文介绍了一个计算斐波那契数列的程序实现，输入为一系列不大于500的整数，输出对应的斐波那契数值。示例展示了输入5、7、11时，输出分别为5、13、89。

1277: Fibonacci Freeze

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	2332	552	Standard

The Fibonacci numbers (0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...) are defined by the recurrence:

Write a program to calculate the Fibonacci Numbers.

Input and Output

The input to your program would be a sequence of numbers smaller or equal than 500, each on a separate line, specifying which Fibonacci number to calculate.

Your program should output the Fibonacci number for each input value, one per line.

Sample Input

5
7
11

Sample Output

The Fibonacci number for 5 is 5
The Fibonacci number for 7 is 13
The Fibonacci number for 11 is 89

This problem is used for contest: 111

Submit / Problem List / Status / Discuss

Problem Set with Online Judge System Version 3.12
Jilin University

Developed by skywind, SIYEE

#include<cstdio>
int a[200];
int b[200];
void add(int *p1,int *p2)
{
    int i;
    for(i=0;p1[i]!=-1&&p2[i]!=-1;i++)
    {
      p1[i]=p1[i]+p2[i];
      if(p1[i]>9)
      {
          p1[i]=p1[i]%10;
          if(p1[i+1]!=-1)
          {
              p1[i+1]++;
          }
          else
          {
              p1[i+1]=1;
              p1[i+2]=-1;
          }
      }
    }
    if(p1[i]==-1&&p2[i]!=-1)
    {
        while(p2[i]!=-1)
        {
            if(p2[i]>9)
            {
                p1[i]=p2[i]%10;
                if(p2[i+1]!=-1)
                {
                    p2[i+1]++;
                }
                else
                {
                    p2[i+1]=1;
                    p2[i+2]=-1;
                }
                i++;
            }
            else
            {
                p1[i]=p2[i];
                i++;
            }
        }
        p1[i]=-1;
    }
    else if(p1[i]!=-1&&p2[i]==-1)
    {
        while(p1[i]!=-1)
        {
            if(p1[i]>9)
            {
                p1[i]=p1[i]%10;
                if(p1[i+1]!=-1)p1[i+1]++;
                else
                {
                    p1[i+1]=1;
                    p1[i+2]=-1;
                }
            }
            i++;
        }
    }
}
int main()
{
    int n;
    int *p;
    while(scanf("%d",&n)==1)
    {
        a[0]=0;a[1]=-1;
        b[0]=1;b[1]=-1;
        for(int i=1;i<n;i++)
        {
            if(i&1)
            {
                p=a;
                add(p,b);
            }
            else
            {
                p=b;
                add(p,a);
            }
        }
        if(n==1||n==0)
        {
            printf("The Fibonacci number for %d is ",n);
            if(n==0)printf("0\n");
            else printf("1\n");
        }
        else
        {
            int len=0;
            while(p[len]!=-1)len++;
            printf("The Fibonacci number for %d is ",n);
            for(int i=len-1;i>=0;i--)printf("%d",p[i]);
            printf("\n");
        }