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hdoj 1250 <10000以内Fibonacci---数组打表>

尤克乔乔 2022-11-21 阅读 34


Hat's Fibonacci


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10566    Accepted Submission(s): 3506


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


戴帽子的



内存狂爆--每个数组最大可以存8个数--


代码:

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int shu[10050][420];
int main()
{
memset(shu,0,sizeof(shu));
shu[1][1]=shu[2][1]=shu[3][1]=shu[4][1]=1;
for (int i=5;i<=10020;i++)
{
for (int j=1;j<=410;j++)
{
shu[i][j]+=shu[i-1][j]+shu[i-2][j]+shu[i-3][j]+shu[i-4][j];
if (shu[i][j]>=10000000)
{
shu[i][j+1]=shu[i][j]/10000000;
shu[i][j]%=10000000;
}
}
}
int n;
while (~scanf("%d",&n))
{
if (n<5)
printf("1\n");
else
{
bool fafe=false;
for (int i=410;i>0;i--)
{
if (fafe)
{
printf("%07d",shu[n][i]);
}
else if (shu[n][i])
{
fafe=true;
printf("%d",shu[n][i]);
}
}
printf("\n");
}
}
return 0;
}



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