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fibonacci高精度加法

时间:2015-08-16 19:38:31      阅读:113      评论:0      收藏:0      [点我收藏+]

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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.

#include <iostream> #include <cstdio> #include <algorithm> #include <cmath> #include <string.h> #include <malloc.h> using namespace std; void add(char* a,char* b,char* c) { int i,j,k,max,min,n,temp; char *s,*pmax,*pmin; max=strlen(a); min=strlen(b); if (max<min) { temp=max; max=min; min=temp; pmax=b; pmin=a; } else { pmax=a; pmin=b; } s=(char*)malloc(sizeof(char)*(max+1)); s[0]=0; for (i=min-1,j=max-1,k=max; i>=0; i--,j--,k--) s[k]=pmin[i]-0+pmax[j]; for (; j>=0; j--,k--) s[k]=pmax[j]; for (i=max; i>=0; i--) if (s[i]>9) { s[i]-=10; s[i-1]++; } if (s[0]==0) { for (i=0; i<=max; i++) c[i-1]=s[i]; c[i-1]=\0; } else { for (i=0; i<=max; i++) c[i]=s[i]; c[i]=\0; } free(s); }
char a[8001][2505]; int main(void) { int i,n; for(i=1;i<=4;i++) strcpy(a[i],"1"); for(i=5;i<8000;i++) { char x[2505],y[2505]; add(a[i-1],a[i-2],x); add(a[i-3],a[i-4],y); add(x,y,a[i]); } while(cin>>n) { cout<<a[n]<<endl; } return 0; }

 

fibonacci高精度加法

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原文地址:http://www.cnblogs.com/chen9510/p/4734700.html

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