hdu 1250 Hat's Fibonacci

http://acm.hdu.edu.cn/showproblem.php?pid=1250

 

Hat's Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3943    Accepted Submission(s): 1346

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.

 

Author

戴帽子的

 

Recommend

Ignatius.L

 

自己写的不怎么样，用了500+MS  不过看到了一个牛B的代码，还是保存下来，有时间慢慢研究。

我的代码：

#include <stdio.h>#include <string.h>#define MAX 3000+3char str1[MAX],str2[MAX],c[MAX];void jia(char str1[],char str2[],char sum[]){    int i,j,k,z;    z=0;    k=0;    for(i=strlen(str1)-1,j=strlen(str2)-1;i>=0||j>=0;i--,j--)    {        if(i>=0)            z+=str1[i]-'0';        if(j>=0)            z+=str2[j]-'0';        c[k++]=z%10+'0';        z/=10;    }    if(z) c[k++]='1';    i=0;    for(--k;k>=0;k--)        sum[i++]=c[k];    sum[i]='\0';}int main(){    int i,n;    //char str1[MAX]="0",str2[MAX]="1",f[MAX];    while(scanf("%d",&n)==1)    {        char str1[MAX]="1",str2[MAX]="1",str3[MAX]="1",str4[MAX]="1",f[MAX],f1[MAX],sum[MAX];        if(n==0)        {            printf("0\n");            continue;        }        for(i=4;i<n;i++)        {            jia(str1,str2,f);            jia(str3,str4,f1);            jia(f,f1,sum);            strcpy(str1,str2);            strcpy(str2,str3);            strcpy(str3,str4);            strcpy(str4,sum);        }        printf("%s\n",str4);    }    return 0;}

http://www.cnblogs.com/newpanderking/archive/2011/07/31/2122528.html

用二维数组模拟大数加法，每一行表述一个数，每一行的一个元素可以代表n位数，这个可以根据自己的需要自己定义。

其他的就和正常的加法一样了，注意进位处理。

代码：

#include <iostream>#include <stdio.h>using namespace std;int s[7500][670];void solve(){    s[1][1] = 1;s[2][1] = 1;    s[3][1] = 1;s[4][1] = 1;    int i,j,k=0;    for(i = 5;i<7500;i++)    for( j = 1;j<=670 ;j++)    {        k += s[i-1][j]+s[i-2][j]+s[i-3][j]+s[i-4][j];        s[i][j] = k%10000;        k = k/10000;    }    while(k)    {        s[i][j++] = k%10000;        k = k/10000;    }}int main(){    int n,i,j;    solve();    while(cin>>n)    {        for(i = 670; i>=1;i--)        if(s[n][i]!=0)break;        printf("%d",s[n][i]);        for(j = i-1;j>=1;j--)        printf("%04d",s[n][j]);        printf("\n");    }}