HDU 1250-Hat's Fibonacci
2013-04-06 22:03
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Hat's Fibonacci
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4730 Accepted Submission(s): 1610
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.
[align=left]Output[/align]
For each case, output the result in a line.
题目给出公式
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),然后给出n,要求出F
,数据比较大,要用到大数运算
AC code:
#include<iostream> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; int f[4][2100]; int ans[2100]; int main() { int N; while(cin>>N) { if(N<=4) { cout<<"1"<<endl; continue; } memset(f,0,sizeof f); memset(ans,0,sizeof ans); f[0][0]=f[1][0]=f[2][0]=f[3][0]=1; int len=1; for(int i=4;i<N;i++) { int carry=0; for(int j=0;j<len;j++) { int temp=carry+f[0][j]+f[1][j]+f[2][j]+f[3][j]; ans[j]=temp%10; carry=temp/10; } while(carry!=0) { ans[len++]=carry%10; carry/=10; } for(int j=0;j<len;j++) { f[i%4][j]=ans[j]; } } for(int i=len-1;i>=0;i--) cout<<ans[i]; cout<<endl; } }
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