hdu 1250 Hat's Fibonacci（大数加法）

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.

 

代码：

#include <iostream>#include<cstdio>#include<cmath>#include<string.h>#define MAXI 7050using namespace std;void bigadd(char a[],char b[],char c[]){    int i,j,la,lb,lc,flag=0,k=0;    char t;    i=strlen(a)-1;    j=strlen(b)-1;    while(i>=0&&j>=0)    {        t=a[i--]+b[j--]+flag-'0';        flag=(t-'0')/10;        t=t-flag*10;        c[k++]=t;    }    while(i>=0)    {        t=a[i--]+flag;        flag=(t-'0')/10;         t=t-flag*10;        c[k++]=t;    }    while(j>=0)    {        t=b[j--]+flag;        flag=(t-'0')/10;         t=t-flag*10;        c[k++]=t;    }    if(flag){c[k]=flag+'0'; c[k+1]='\0';}    else{ k--;c[k+1]='\0';}    for(i=0,j=strlen(c)-1;i<j;i++,j--)    {        t=c[i];c[i]=c[j];c[j]=t;    } }char a[MAXI][2050],c[2050],d[2050];int main(){   int i,j,n;   strcpy(a[1],"1");   strcpy(a[2],"1");   strcpy(a[3],"1");   strcpy(a[4],"1");   for(i=5;i<MAXI;i++)   {   bigadd(a[i-1],a[i-2],c);   bigadd(c,a[i-3],d);   bigadd(d,a[i-4],a[i]);   }       while(~scanf("%d",&n))   {   printf("%s\n",a[n]);}return 0;}