HDU1250(大数相加)

发布时间：2020-12-14 04:10:26 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): 4998????Accepted Submission(s): 1691 Problem Description A Fibonacci sequence is calculated by adding the previous two

Hat's Fibonacci

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4998????Accepted Submission(s): 1691

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

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.

///运用数组存储固定邻接位数

#include<iostream> #include<iomanip> using namespace std; struct { ?int link[600]; ?int num; }a[10010]; int main() { ?int n,i,j,temp; ?a[1].link[a[1].num++]=1; ?a[2].link[a[2].num++]=1; ?a[3].link[a[3].num++]=1; ?a[4].link[a[4].num++]=1; ?for(i=5;i<=10000;i++) ?{ ??temp=0; ??for(j=0;j<=a[i-1].num-1;j++) ??{ ???a[i].link[j]=a[i-1].link[j]+a[i-2].link[j]+a[i-3].link[j]+a[i-4].link[j]+temp; ???temp=a[i].link[j]/10000; ???a[i].link[j]=a[i].link[j]%10000; ??} ??a[i].num=a[i-1].num; ??if(temp>0) ??{ ???a[i].link[j]=temp; ???a[i].num++; ??} ?} ?while(cin>>n) ?{ ??j=a[n].num-1; ??cout<<a[n].link[j]; ??for(j=a[n].num-2;j>=0;j--) ??{ ???printf("%04d",a[n].link[j]);?? ??} ??cout<<endl;? ?} ?return 0; }

（编辑：李大同）

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