Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 11104 Accepted Submission(s): 3732
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
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1250
分析:这题用到大数相加,用数组的元素表示大数的各个数位的数字,(例如123,可以a[0]=3,a[1]=2,a[2]=1;)有个技巧是在网上学到的,每个数组元素存储八位数可以提高效率。先预处理,再输入数据。
下面给出AC代码:
1 #include<bits/stdc++.h>
2 using namespace std;
3 int a[10000][260]={0}; //每个元素可以存储8位数字,所以2005位可以用260个数组元素存储。
4 int main()
5 {
6 int i,j,n;
7 a[1][0]=1; //赋初值
8 a[2][0]=1;
9 a[3][0]=1;
10 a[4][0]=1;
11 for(i=5;i<10000;i++)
12 {
13 for(j=0;j<260;j++)
14 a[i][j]+=a[i-1][j]+a[i-2][j]+a[i-3][j]+a[i-4][j];
15 for(j=0;j<260;j++) //每八位考虑进位。
16 if(a[i][j]>100000000)
17 {
18 a[i][j+1]+=a[i][j]/100000000;
19 a[i][j]=a[i][j]%100000000;
20 }
21 }
22 while(scanf("%d",&n)!=EOF)
23 {
24 for(j=259;j>=0;j--)
25 if(a[n][j]!=0) break; //不输出高位的0
26 printf("%d",a[n][j]);
27 for(j=j-1;j>=0;j--)
28 printf("%08d",a[n][j]); //每个元素存储了八位数字,所以控制输出位数为8,左边补0
29 printf("\n");
30 }
31 return 0;
32 }