杭电1250Hat's Fibonacci
发布时间：2020-12-14 02:25:57 所属栏目：大数据 来源：网络整理
导读：Hat's Fibonacci Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9134????Accepted Submission(s): 2970 Problem Description A Fibonacci sequence is calculated by adding the previous two
Hat's FibonacciTime Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9134????Accepted Submission(s): 2970
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.
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Input
Each line will contain an integers. Process to end of file.
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Output
For each case,output the result in a line.
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Sample Input
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Sample Output
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Author 妥妥的水题，但是要是存二维数组，第二维只存一位数的话会超内存的，超了3次才想到存4位数；所以第二维只需要开到2005/4即可；一维开到8000即可.
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define N 2010
int a[7061][555];
int i,n,m,j,k,l;
int main()
{
memset(a,sizeof(a));
a[1][0]=a[2][0]=a[3][0]=a[0][0]=1;
for(i=4;i<7060;i++)
{
k=0;
for(j=0;j<555;j++)
{
a[i][j]=a[i-1][j]+a[i-2][j]+a[i-3][j]+a[i-4][j]+k;
k=a[i][j]/10000;
a[i][j]=a[i][j]%10000;
}
}
while(scanf("%d",&n)!=EOF)
{
j=550;
while(a[n-1][j]==0)
j--;
printf("%d",a[n-1][j--]);//这一位不一定是四位数，所以输出%d
for(; j>=0; j--)
printf("%04d",a[n-1][j]);//剩下的一定是4位数的，所以输出%04d，否则就会错.！！！！！！
putchar('n');
//printf("%d",a[n-1][j]);//这的输出也是ac代码；
//i=j-1;//个人比较喜欢下边的输出
//for(i=j-1;i>=0;i--)
//printf("%04d",a[n-1][i]);
//printf("n");
}
return 0;
}
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