HDU 1250 大数加斐波那契数列

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

/*
练习简单DP是碰到的一道题，更像是大数相加
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
*/
#include <stdio.h>
#include <string.h>

int data[7250][600];
int p;
int main()
{
    memset(data,0,sizeof(data));
    data[1][0]=1;
    data[2][0]=1;
    data[3][0]=1;
    data[4][0]=1;
    p=1;//各数种最高位所在加1
    for(int i=5;i<=7200;i++)//循环计算各数
    {
        for(int j=0;j<=p;j++)
           data[i][j]=data[i-1][j]+data[i-2][j]+data[i-3][j]+data[i-4][j];
        for(int j=0;j<=p;j++)
        {
            data[i][j+1]+=data[i][j]/10000;
            data[i][j]%=10000;
        }
/*
        printf("\n%d  \n",i);
        for(int j=p;j>=0;j--)
            printf("%d",data[i][j]);
*/
        if(data[i][p])
            p++;
    }

    int n;
    //printf("p=%d\n",p);//衡量2003位数需要的i值
    while(scanf("%d",&n)!=EOF)
    {

        int i;

        for(i=p;i>=0;i--)
            if(data[n][i]!=0)
                break;
        printf("%d",data[n][i--]);//第一位不需要前置0
       for(;i>=0;i--)
       {
           printf("%04d",data[n][i]);//不足4位前置0
       }

        printf("\n");
    }
    return 0;
}