Hat's Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4400    Accepted Submission(s): 1480

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.

 

先声明一个数组data，前四个数置1，后用大数加法实现4个数相加并打表，注意每位存放10000，一开始存放10,结果超内存

1 /* 2 练习简单DP是碰到的一道题，更像是大数相加 3 先声明一个数组data，前四个数置1，后用大数加法实现4个数相加并打表， 4 注意每位存放10000，一开始存放10,结果超内存 5 */ 6 #include 
     
       7 #include 
      
        8  9 int data[7250][600];10 int p;11 int main()12 {13     memset(data,0,sizeof(data));14     data[1][0]=1;15     data[2][0]=1;16     data[3][0]=1;17     data[4][0]=1;18     p=1;//各数种最高位所在加119     for(int i=5;i<=7200;i++)//循环计算各数20     {21         for(int j=0;j<=p;j++)22            data[i][j]=data[i-1][j]+data[i-2][j]+data[i-3][j]+data[i-4][j];23         for(int j=0;j<=p;j++)24         {25             data[i][j+1]+=data[i][j]/10000;26             data[i][j]%=10000;27         }28 /*29         printf("\n%d  \n",i);30         for(int j=p;j>=0;j--)31             printf("%d",data[i][j]);32 */33         if(data[i][p])34             p++;35     }36 37     int n;38     //printf("p=%d\n",p);//衡量2003位数需要的i值39     while(scanf("%d",&n)!=EOF)40     {41 42         int i;43 44         for(i=p;i>=0;i--)45             if(data[n][i]!=0)46                 break;47         printf("%d",data[n][i--]);//第一位不需要前置048        for(;i>=0;i--)49        {50            printf("%04d",data[n][i]);//不足4位前置051        }52 53         printf("\n");54     }55     return 0;56 }