fibonacci高精度加法

最新推荐文章于 2018-05-07 18:06:59 发布

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最新推荐文章于 2018-05-07 18:06:59 发布

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原文链接：http://www.cnblogs.com/chen9510/p/4734700.html

本文介绍了一个计算四阶斐波那契数的C++程序实现，该程序能够处理大整数输入并输出相应的斐波那契数值。通过递推公式计算，并使用字符串进行大数运算。

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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

100

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.


#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string.h>
#include <malloc.h>
using namespace std;

void add(char* a,char* b,char* c)
{
    int i,j,k,max,min,n,temp;
    char *s,*pmax,*pmin;
    max=strlen(a);
    min=strlen(b);
    if (max<min)
    {
        temp=max;
        max=min;
        min=temp;
        pmax=b;
        pmin=a;
    }
    else
    {
        pmax=a;
        pmin=b;
    }
    s=(char*)malloc(sizeof(char)*(max+1));
    s[0]='0';
    for (i=min-1,j=max-1,k=max; i>=0; i--,j--,k--)
        s[k]=pmin[i]-'0'+pmax[j];
    for (; j>=0; j--,k--)
        s[k]=pmax[j];
    for (i=max; i>=0; i--)
        if (s[i]>'9')
        {
            s[i]-=10;
            s[i-1]++;
        }
    if (s[0]=='0')
    {
        for (i=0; i<=max; i++)
            c[i-1]=s[i];
        c[i-1]='\0';
    }
    else
    {
        for (i=0; i<=max; i++)
            c[i]=s[i];
        c[i]='\0';
    }
    free(s);
}
char a[8001][2505];
int main(void)
{
    int i,n;
    for(i=1;i<=4;i++)
    strcpy(a[i],"1");
    for(i=5;i<8000;i++)
    {
        char x[2505],y[2505];
        add(a[i-1],a[i-2],x);
        add(a[i-3],a[i-4],y);
        add(x,y,a[i]);
    }
    while(cin>>n)
    {
        cout<<a[n]<<endl;
    }
    return 0;
}