输出斐波那契数前10000项

来源：互联网发布：软件评测师报名编辑：程序博客网时间：2024/06/05 20:09

之前也写过输出斐波那契数的程序，当然也只是前几十位。所以也没注意到当输出大于47位之后，就发生溢出了。直到某一天，老师让我们输出47位之后的数我才发现这个问题。于是就花了一点时间写了出来。

其实这个问题归根到底就是一个大数相加的问题，当然网上也能搜到很多关于大于数加减乘除的模版。我现在就换另外一种方法来计算大数相加的问题。思路就是分别用两个字符串来保存两个数，然后对应位分别相加，满10进1，就行了。理论上是没有上限的（不会溢出），但受计算机的限制，在我自己的电脑上也只能输出几万项斐波那契数。不多说，上代码。

//定义了一个大数类

#ifndef BIGNUM_H#define BIGNUM_H#include <iostream>using namespace std;class BigNum{private:    string num;    int length;public:    BigNum()        :num("NULL")        ,length(0)    {}    BigNum(string &n)        :num(n)        ,length(n.size())    {}    BigNum(string n,int l)        :num(n)        ,length(l)    {}    void getNum() const;    const void clear();    const BigNum &operator +(const BigNum & n);    const BigNum &operator =(const BigNum & n);};#endif // BIGNUM_H

#include "bignum.h"using namespace std;void BigNum::getNum()const{    cout<<num<<" "<<length<<endl;}const void BigNum::clear(){    num.clear();    length = 0;}const BigNum &BigNum::operator =(const BigNum & n){    num = n.num;    length = n.length;    return *this;}const BigNum &BigNum::operator +(const BigNum & n){    //      num += n.num;    //      length +=n.length;    //      return *this;    BigNum maxNum = n ,minNum = *this,sumNum;    sumNum.num.clear();    int temp;    int reamain = 0;    if(length > n.length)    {        maxNum = *this;        minNum = n;    }  //  cout<<endl; cout<<endl;  //  maxNum.getNum();   // minNum.getNum();  //  cout<<endl;    int maxLen = maxNum.length - 1,minLen = minNum.length - 1;    for(int i = 0; i < minNum.length; i++)    {        temp = (maxNum.num[maxLen] - '0') + (minNum.num[minLen] - '0') + reamain;        sumNum.num += (temp % 10 + '0');        reamain = temp / 10;        maxLen --;        minLen --;        sumNum.length ++;       // cout<<"r1 = "<<reamain<<"  ";        // sumNum.getNum();    }    //cout<<" maxLen = "<<maxLen<<      //    " minLen = "<<minLen<<       //   " reamain = "<<reamain<<       //   " temp = "<<temp<<endl;   // cout<<"---------"<<endl;    for(int j = maxLen; j >= 0; j--)    {        temp = maxNum.num[j] - '0' + reamain;        sumNum.num += temp % 10 + '0';        reamain = temp / 10 ;        sumNum.length ++;        //cout<<"r2 = "<<reamain<<"  ";        // sumNum.getNum();    }    //cout<<endl<<"---------"<<endl;    if(reamain != 0)    {        sumNum.num += (reamain + '0');        sumNum.length++;       // cout<<"r3 = "<<reamain<<"  ";    }   // cout<<endl;    string s(sumNum.num.rbegin(),sumNum.num.rend());    num = s;    length = sumNum.length;    return *this;}

#include <iostream>#include "bignum.h"using namespace std;void Swap(BigNum &a, BigNum &b){    BigNum temp;    temp = a;    a = b;    b = temp;}const BigNum &Fibonacci(int n, BigNum &temp){    BigNum a("1",1);    BigNum b("1",1);    temp.clear();    // a.getNum();    // cout<<"==========";    for(int i = 3; i <= n; i++)    {        a + b;        temp = a;        Swap(a,b);        //  cout<<"a = ";        //  a.getNum();        // cout<<"b = ";        //b.getNum();    }    // cout<<"=========="<<endl;    return temp;}int main(){    int num;    BigNum f;    for(;;)    {        cout<<"input:";        cin>>num;        cout<<endl;        if(num == 1 || num == 2)        {            cout<<"the num is: "<<1<<endl;        }        else        {            f = Fibonacci(num,f);            cout<<"the num is: ";            f.getNum();            cout<<endl;        }    }    return 0;}

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