hdu 1250 高精度+类似斐波那契数列

最新推荐文章于 2024-08-26 15:30:00 发布

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最新推荐文章于 2024-08-26 15:30:00 发布

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分类专栏：高精度

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本文介绍了一种使用高精度算法计算Fibonacci数的方法，通过存储和处理长整型数字来解决传统数据类型无法表示大数值的问题。该算法采用逐位相加的方式，并考虑进位情况，能够有效计算出任意指定位置的Fibonacci数。

http://acm.hdu.edu.cn/showproblem.php?pid=1250

Hat's Fibonacci

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7245 Accepted Submission(s): 2368

Problem Description

A Fibonacci sequence is calculated by adding the previous two members the sequence, with the first two members being both 1.
F(1) = 1, F(2) = 1, F(3) = 1,F(4) = 1, F(n>4) = F(n - 1) + F(n-2) + F(n-3) + F(n-4)
Your task is to take a number as input, and print that Fibonacci number.

Input

Each line will contain an integers. Process to end of file.

Output

For each case, output the result in a line.

Sample Input

Sample Output

  
   4203968145672990846840663646


Note:
No generated Fibonacci number in excess of 2005 digits will be in the test data, ie. F(20) = 66526 has 5 digits.

就是一个简单的高精度水题

#include <string.h>
#include <stdio.h>
#include <iostream>
#define M 2010
using namespace std;
char data[10000][M+2];
int main()
{
    int i,j,p;
    memset(data,0,sizeof(data));
    data[1][2010]=1;
    data[4][2010]=1;
    data[2][2010]=1;
    data[3][2010]=1;
    i=5;p=2010;
    //int count=0;
    while(data[i-1][5]<=1)
    {
        for(j=2010;j>=p;j--)
        {
            data[i][j]=data[i-1][j]+data[i-2][j]+data[i-3][j]+data[i-4][j];
        }
        for(j=2010;j>=p;j--)
        {
            int c=data[i][j]/10;
            if(c>=1)
            {
                data[i][j]=data[i][j]%10;
                data[i][j-1]=data[i][j-1]+c;
            }
        }
        if(data[i][p-1]>0)
            p--;
        i++;
        //count++;
    }
    //printf("%d\n",count);
    int n;
    while(~scanf("%d",&n))
    {

        int temp;
        for(int i=0;i<=2010;i++)
            if(data[n][i]+'0'!='0')
            {
                temp=i;
                break;
            }
        for(int i=temp;i<=2010;i++)
            printf("%c",data[n][i]+'0');
        printf("\n");
    }
    return 0;
}