(二维数组 亿进制 或 滚动数组） Hat's Fibonaccihdu1250

Hat's Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 12284    Accepted Submission(s): 4124

 

 

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.

 

用string会超时，以下为超时代码。

#include <iostream> #include <string> using namespace std; string add(string a,string b) { int len1=a.length(); int len2=b.length(); int i; if(len1>len2) { for(i=1;i<=len1-len2;i++) b="0"+b; } else { for(i=1;i<=len2-len1;i++) a="0"+a; } string str; int cf=0,t; len1=a.length(); for(i=len1-1;i>=0;i--) { t=a[i]-'0'+b[i]-'0'+cf; cf=t/10; t%=10; str=char(t+'0')+str; } if(cf!=0) str=char(cf+'0')+str; return str; } string fun(int n) { string f[10010]; f[1]="1"; f[2]="1"; f[3]="1"; f[4]="1"; int i; string a,b,c; for(i=5;i<=n;i++) { a=add(f[i-1],f[i-2]); b=add(f[i-3],f[i-4]); f[i]=add(a,b); } return f[n]; } int main() { int n; while(cin>>n) { string str; str=fun(n); cout<<str<<endl; cout<<endl; } return 0; } View Code

 正确的代码

方法一： 利用二维数组和亿进制。 #include<cstdio> #include <iostream> #include<cstring> using namespace std; int str[10001][260]; //必须为int类型。 int main() { memset(str,0,sizeof(str)); str[1][0]=1; str[2][0]=1; str[3][0]=1; str[4][0]=1; int i,j,ans=0,c,n; for(i=5;i<10001;i++) { for(j=0,c=0;j<260;j++) //循环，来将j个数组的8位数字的情况全部列举出。 { ans=str[i-1][j]+str[i-2][j]+str[i-3][j]+str[i-4][j]+c; c=ans/100000000; str[i][j]=ans%100000000; //每一个数组存8位数字，c来判定是否进位。 } } while(cin>>n) { j=259; while(!str[n][j]) //首位有0，清除掉0。 j--; cout<<str[n][j]; //开头的首0清除掉后的x位数字，可能小于8位。 for(i=j-1;i>=0;i--) printf("d",str[n][i]); //每8位数字输出一组，不足的自动部0。 printf("\n"); } return 0; } 方法二： 利用滚动数组求解 #include <iostream> #include <cstdio> #include <cstring> using namespace std; int t[6][2555]; int main() { int n; while(scanf("%d",&n) != EOF) { memset(t,0,sizeof(t)); t[0][0] = 1; t[1][0] = 1; t[2][0] = 1; t[3][0] = 1; for(int i = 4;i < n;i++) { int carry = 0; int k = i % 4; for(int j = 0;j < 2500;j++) { int x = t[0][j] + t[1][j] + t[2][j] + t[3][j]; t[k][j] = x + carry; carry = t[k][j] / 10; t[k][j] %= 10; } } int k = 2500; while(t[(n - 1) % 4][--k] == 0); for(int i = k;i >= 0;i--) { printf("%d",t[(n - 1) % 4][i]); } printf("\n"); } return 0; } View Code

 

 用JAVA

import java.math.BigInteger; import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner in = new Scanner (System.in); BigInteger a[]=new BigInteger[10001]; int n; while(in.hasNextInt()) { n=in.nextInt(); a[1]=BigInteger.ONE; a[2]=BigInteger.ONE; a[3]=BigInteger.ONE; a[4]=BigInteger.ONE; for(int i=5;i<=10000;i++) { a[i]=BigInteger.ZERO; a[i]=a[i].add(a[i-1]); a[i]=a[i].add(a[i-2]); a[i]=a[i].add(a[i-3]); a[i]=a[i].add(a[i-4]); } System.out.println(a[n]); } } } View Code

 

转载于:https://www.cnblogs.com/Weixu-Liu/p/9165370.html

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