hust 1003 - Sibonacci Numbers

最新推荐文章于 2015-01-26 00:25:49 发布

cinyaaaa

最新推荐文章于 2015-01-26 00:25:49 发布

阅读量766

点赞数

CC 4.0 BY-SA版权

分类专栏： hust 机智小水题文章标签： hust 1003 - Sibonacc

本文链接：https://blog.youkuaiyun.com/cxy450019566/article/details/42581981

机智小水题同时被 2 个专栏收录

11 篇文章

订阅专栏

hust

4 篇文章

订阅专栏

本文介绍了Sibonacci数列的概念及其与经典Fibonacci数列的区别，详细解释了Sibonacci数列的定义和计算方法，并提供了一段C语言实现的代码样例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

原题链接：

http://acm.hust.edu.cn/problem/show/1003

1003 - Sibonacci Numbers

Time Limit: 1s Memory Limit: 64MB

Submissions: 2029 Solved: 348

DESCRIPTION

As is known to all, the definition of Fibonacci Numbers is: f(1)=1 f(2)=1 f(n)=f(n-1)+f(n-2) (n>=3) Now Sempr found another Numbers, he named it "Sibonacci Numbers", the definition is below: f(x)=0 (x<0) f(x)=1 (0<=x<1) f(x)=f(x-1)+f(x-3.14) (x>=1) Your work is to tell me the result of f(x), is the answer is too large, divide it by 1000000007 and give me the remainder. Be careful the number x can be an integer or not.

INPUT

In the first line there is an Integer T(0<T<10000) which means the number of test cases in the input file. Then followed T different lines, each contains a number x(-1000<x<1000).

OUTPUT

For each case of the input file, just output the result, one for each line.

SAMPLE INPUT

SAMPLE OUTPUT

0
1
2

HINT

SOURCE

Sempr|CrazyBird|hust07p43

代码：

#include "stdio.h"
#include "string.h"
#include "math.h"

int f[2000010],mod=1000000007;
int xx;
double x;

long long getf(int a)
{
	if(a<0)
		return 0;
	if(a<1000)
	{
		return 1;
	}
	if(f[a]!=-1)
		return f[a];
	else
	{
		return (getf(a-1000)+getf(a-3140))%mod;
	}
}

void solve()
{
	for(int i=1;i<=1000000;i++)
	{
		f[i]=getf(i);
	}
}

int main()
{
	int T;
	scanf("%d",&T);
	memset(f,-1,sizeof(f));
	solve();
	while(T--)
	{
		scanf("%lf",&x);
		x=floor(x*1000);
		xx=x;
		printf("%d\n",getf(xx));
	}
	return 0;
}