ZOJ 1828 Fibonacci Numbers

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.

Sample Input

100

Sample Output

354224848179261915075

Note:

No generated Fibonacci number in excess of 1000 digits will be in the test data,i.e. f(20) = 6765 has 4 digits.

分析：数据太大，模拟大数加法

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
int f[5005][1010];		//保存最终结果  
int len[5005];			//保存最终结果的长度  

void display()
{
	f[1][1] = 1;
	f[2][1] = 1;
	len[1] = 1;
	len[2] = 1;
	int i,j;
	for (i = 3; i <= 5000; i++)
	{
		int jin = 0;	//判断是否有进位  
		for (j = 1; j <= max(len[i - 1],len[i - 2]) ; j++)
		{
			f[i][j] = f[i - 1][j] + f[i - 2][j] + jin;
			jin = f[i][j] / 10;
			f[i][j] %= 10;
		}
		if (jin)
		{
			f[i][j] = 1;
			len[i] = j;
		}
		else 
		{
			len[i] = j - 1;
		}
	}
}
int main()
{
	display();
	int n;
	while (~scanf("%d",&n))
	{
		for (int i = len[n]; i >= 1; i--)
			printf("%d",f[n][i]);
		printf("n");
	}
	return 0;
}