hdu1250 Hat's Fibonacci

Hat's Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12633    Accepted Submission(s): 4241

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

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

 

Author

//题目中说答案中的结果不会超过2005位数字，而我一个int 存了8位，所以可以确定数组的第二维
//最多开个260 260*8=2080>2005就行，而1维嘛，10的2006次方大约等于2的7000多次方，所以开个8000足够； 
#include
int i,j,n;
int ans[8000][255]={{0}};//先打表再查表 ans[个数][这个数的每一位的值] 
void fun()
{
	for(i=1;i<5;i++)
	ans[i][1]=1;//注意，这里为了方便，二维都是从1开始的 
	
	for(i=5;i<8000;i++)
	for(j=1;j<255;j++)
	{
		ans[i][j]+=ans[i-1][j]+ans[i-2][j]+ans[i-3][j]+ans[i-4][j];
    	ans[i][j+1]+=ans[i][j]/100000000;
    	ans[i][j]%=100000000;
	}
}

int main()
{  

	fun();
	while(scanf("%d",&n)!=EOF)
	{
		for(i=254;i>0;i--)//从前往后找到有效值 
			if(ans[n][i])
				break;
		printf("%d",ans[n][i]);//为了不输出前导0，所以这里特别输出 
		i--; 
		for(i;i>0;i--)//继续使用上面循环中的i,并减1 
	 	printf("%.8d",ans[n][i]);
		printf("\n");
	}
	return 0;
}