Hat's Fibonacci（大数）

 

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

100

 

Sample Output

4203968145672990846840663646Note:No generated Fibonacci number in excess of 2005 digits will be in the test data, ie. F(20) = 66526 has 5 digits.

 

Author

戴帽子的

 

 

Recommend

Ignatius.L

题目给出公式 

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)，然后给出n，要求出F[n]，数据比较大，要用到大数运算 

代码：

#include<iostream>#include<cstdlib>#include<cstring>#include<algorithm>#include<cstdio>using namespace std;int f[4][2101];int ans[2101];int main(){    int N;    while(cin>>N)    {        if(N<=4)        {            cout<<"1"<<endl;            continue;        }        memset(f,0,sizeof (f));        memset(ans,0,sizeof (ans));        f[0][0]=f[1][0]=f[2][0]=f[3][0]=1;        int len=1;        for(int i=4;i<N;i++)        {            int carry=0;            for(int j=0;j<len;j++)            {                int temp=carry+f[0][j]+f[1][j]+f[2][j]+f[3][j];                ans[j]=temp%10;                carry=temp/10;            }            while(carry!=0)            {                ans[len++]=carry%10;                carry/=10;            }            for(int j=0;j<len;j++)            {                f[i%4][j]=ans[j];            }        }        for(int i=len-1;i>=0;i--)            cout<<ans[i];        cout<<endl;    }}

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