HDU 1250 Hat's Fibonacci



Hat's Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 8372    Accepted Submission(s): 2725


[align=left]Problem Description[/align]
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.

 

[align=left]Input[/align]
Each line will contain an integers. Process to end of file.

 

[align=left]Output[/align]
For each case, output the result in a line.
 

[align=left]Sample Input[/align]

100

 

[align=left]Sample Output[/align]

4203968145672990846840663646

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

 

[align=left]Author[/align]
戴帽子的

滚动数组+大数。
开个二维数组，第一维是N，第二维是数字个数。。

#include <stdio.h>
#include <algorithm>
#include<iostream>
#include<cstring>
using namespace std;
#include <string.h>
int main()
{
int x[5][2008]; //之所以开5，因为下面所有的操作都MOD 5 了。
int n,i,j;
while(scanf("%d",&n)!=EOF)
{
if(n<=4)
{
printf("1\n");
continue;
}
memset(x,0,sizeof(x));
x[1][0]=1;
x[2][0]=1;
x[3][0]=1;
x[4][0]=1;      //初始化
for(i=5;i<=n;i++)
{
memset(x[i%5],0,sizeof(x[i%5]));  //初始化。
for(j=0;j<=2005;j++)
{
x[i%5][j]+=x[(i-1)%5][j]+x[(i-2)%5][j]+x[(i-3)%5][j]+x[(i-4)%5][j];//大数思想+滚动数组
if(x[i%5][j]>=10)
{
x[i%5][j+1]+=x[i%5][j]/10;   //因为是反过来输出，所以要后一位进位。
x[i%5][j]%=10;
}
}
}
i=2005;
while(x[n%5][i]==0)//从后面开始判断非0。
--i;
while(i>=0)
{
printf("%d",x[n%5][i]); //一个数一个数输出。
--i;
}
printf("\n");
}
return 0;
}