Hat's Fibonacci
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.
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
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 }
