1041: 二哥的困惑 Ⅳ
Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 284 Solved: 91
[Submit][Status][Web Board]
Description
A Fibonacci sequence is calculated by adding the previous two members of the sequence, with the first two members being both 1.
f(1) = 1, f(2) = 1, f(n > 2) = f(n - 1) + f(n - 2)
Your task is to take a number as input, and print that Fibonacci number.
Input
Input contains several cases, every case contains a positive integer number N.
Output
print the Nth Fibonacci number.
Sample Input
100
Sample Output
354224848179261915075
【分析】
下面这个用的是矩阵做的，但是100太大了不行；
#include<bits/stdc++.h>
using namespace std;void cal(long long a[1][2],long long b[2][2])
{long long c[1][2];c[0][0]=a[0][0]*b[0][0]+a[0][1]*b[1][0];c[0][1]=a[0][0]*b[0][1]+a[0][1]*b[1][1];a[0][0]=c[0][0];a[0][1]=c[0][1];
}
int main()
{long long n,cnt=0;while(~scanf("%lld",&n)){long long a[1][2]={1,1};long long b[2][2]={0,1,1,1};if(n==1||n==2)cout<<"1\n";else {for(int i=3;i<=n;i++)cal(a,b);cout<<a[0][1]<<endl;}}
}所以，，，还是用Java吧
import java.math.BigInteger;
import java.util.Scanner;
public class Main {public static void main(String args[]) {Scanner in=new Scanner(System.in);int n;BigInteger a[]=new BigInteger[1005];while(in.hasNext()) {n=in.nextInt();a[1]=BigInteger.valueOf(1);a[2]=BigInteger.valueOf(1);for(int i=3;i<=n;i++)a[i]=a[i-1].add(a[i-2]);System.out.println(a[n]);}}
}关于Java大数类的一些知识点：https://www.cnblogs.com/Leonard-/p/7636639.html
https://www.cnblogs.com/superxuezhazha/p/4761854.html