Problem Description

2007年到来了。经过2006年一年的修炼，数学神童zouyu终于把0到100000000的Fibonacci数列 (f[0]=0,f[1]=1;f[i] = f[i-1]+f[i-2](i>=2))的值全部给背了下来。 接下来，CodeStar决定要考考他，于是每问他一个数字，他就要把答案说出来，不过有的数字太长了。所以规定超过4位的只要说出前4位就可以了，可是CodeStar自己又记不住。于是他决定编写一个程序来测验zouyu说的是否正确。

 

Input

输入若干数字n(0 <= n <= 100000000)，每个数字一行。读到文件尾。

 

Output

输出f[n]的前4个数字（若不足4个数字，就全部输出）。

 

Sample Input

 

Sample Output

 

Author

daringQQ

#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<cmath>

#include<vector>

using namespace std;

const int INF=0x3f3f3f3f;

#define mem(x,y) memset(x,y,sizeof(x))

#define SI(x) scanf("%d",&x)

#define PI(x) printf("%d",x)

#define SD(x,y) scanf("%lf%lf",&x,&y)

#define P_ printf(" ")

int main(){

	int n;

	int dp[20]={0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181};

	while(~SI(n)){

		double ans=log10(1/sqrt(5.0))+n*log10((1+sqrt(5.0))/2.0);

		ans=ans-floor(ans);

		ans=pow(10,ans);

		if(n<20)printf("%d\n",dp[n]);

		else printf("%d\n",(int)(ans*1000));

	}

	return 0;

}

#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<cmath>

#include<vector>

using namespace std;

const int INF=0x3f3f3f3f;

#define mem(x,y) memset(x,y,sizeof(x))

#define SI(x) scanf("%d",&x)

#define PI(x) printf("%d",x)

#define SD(x,y) scanf("%lf%lf",&x,&y)

#define P_ printf(" ")

const int MAXN=100010;

char c[MAXN];

int x[MAXN],y[MAXN],z[MAXN];

void bigsum(char *a,char *b){

	int len1=strlen(a),len2=strlen(b);

	int len=max(len1,len2);

	mem(x,0);mem(y,0);mem(z,0);

	for(int i=0,j=len1-1;i<len1;i++,j--)x[i]=a[j]-'0';

	for(int i=0,j=len2-1;i<len2;i++,j--)y[i]=b[j]-'0';

	for(int i=0;i<len;i++){

		z[i]=x[i]+y[i]+z[i];

		z[i+1]+=z[i]/10;

		z[i]%=10;

		if(z[len])len++;

	}

	for(int i=0,j=len-1;i<len;i++,j--)c[i]=z[j]+'0';

	c[len]='\0';

}

int main(){

	int n;

	char a[MAXN],b[MAXN];

	while(~SI(n)){

		if(n==0){

			puts("0");continue;

		}

		if(n==1){

			puts("1");continue;

		}

		if(n==2){

			puts("1");continue;

		}

			a[0]='1';a[1]='\0';

			c[0]='2',c[1]='\0';

		for(int i=3;i<n;i++){

			//puts(c);

			strcpy(b,c);

			bigsum(a,c);

			strcpy(a,b);

		}

		for(int i=0;i<4&&c[i];i++)printf("%c",c[i]);puts("");

	}

	return 0;

}

package com.lanqiao.week1;

import java.util.Scanner;

public class hdoj1568 {

    private static Scanner cin;

    private static int MOD = 1000000007;

    static{

        cin = new Scanner(System.in);

    }

    public static void main(String[] args) {

        int n;

        int[] dp = {0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 1094, 1771, 2865};

        //2584 4181 6765 10946 17711 28657

//        for(int i = 0; i < 10; i++){

//            System.out.print(" " + i + "-->" + dp[i]);

//        }

        while(cin.hasNext()){

            n = cin.nextInt();

            //System.out.println(n);

            if(n < dp.length){

                System.out.println(dp[n]);

            }else{

                double x =

                        Math.log10(1.0 / Math.sqrt(5.0))

                        + n * Math.log10((1.0 + Math.sqrt(5.0)) / 2.0);

                x = x - Math.floor(x);

                System.out.println((int)(Math.pow(10, x) * 1000));

            }

        }

    }

}